二维数组中找最大特征数组

atlandisebr 2021-04-27 01:11:55
有一个二维数组a[5000][5000];
a[0]={0,0,0,1,0,0,0,0,1,1,*******0,1,1,0};
a[1]={0,0,1,1,0,1,0,0,0,0,*******1,0,1,0};
*********
a[4998]={1,0,0,1,0,1,0,1,0,0,*******0,0,0,0};
a[4999]={0,1,1,1,0,1,0,0,0,0,*******0,0,1,1};
元素的排列是随机的，a
  • 中1的个数，位置也是随机的；
    需要找到一个b
  • [8],使得符合((((a[*][b[0]]=1)&&(a[*][b[1]]=1))&&((a[*][b[2]]=1)&&(a[*][b[3]]=1)))&&(((a[*][b[4]]=1)&&(a[*][b[5]]=1))&&((a[*][b[6]]=1)&&(a[*][b[7]]=1))))这个条件的a
  • 的个数是最多的，这样的算法应该是怎么样的？望赐教！
    使用过两个FOR循环后+加5000中选8个的组合算法，但是消耗时间太久了，有没有什么快速的办法？
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千梦一生 2021-05-11
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51节前看到的，有个思路，趁机想把代码写下来。 可惜写了一点点下班了，摸鱼党用不加班。接着放51就没继续写了： 想法是： 010111101010100 111111100000111 011101010101011 101110100011110 011111110111011 找到1最多的前8列。设这8列的编号即为b[8]，当然，结果并不一定如此。获得此时这8列中1111 1111的数量n 我的想法是：如果这n值求得了。那么可以在后续的运算中排除1的数量小于n的列。因为这些列不可能存在b[8]中。 我觉得这个思路在特别多列的时候，又是纯随机的数据，效率提升会非常高【比老老实实算下去】
千梦一生 2021-05-11
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51节前看到的，有个思路，趁机想把代码写下来。 可惜写了一点点下班了，摸鱼党用不加班。接着放51就没继续写了： 想法是： 010111101010100 111111100000111 011101010101011 101110100011110 011111110111011 找到1最多的前8列。设这8列的编号即为b[8]，当然，结果并不一定如此。获得此时这8列中1111 1111的数量n 我的想法是：如果这n值求得了。那么可以在后续的运算中排除1的数量小于n的列。因为这些列不可能存在b[8]中。 我觉得这个思路在特别多列的时候，又是纯随机的数据，效率提升会非常高【比老老实实算下去】
atlandisebr 2021-05-07
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引用 96 楼 早打大打打核战争 的回复:
可以考虑把每行看作一个二进制数字，对按行排序过的矩阵再做处理？这样可能不需要C(5000, 8)次循环了，只需要循环5000次，每次处理5000、4999... 1行数据
具体怎么实现呢？
atlandisebr 2021-05-06
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是最多数量的行
GKatHere 2021-05-06
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b[8]的值不重复，找到第八个最多数量的列不就可以了？
atlandisebr 2021-05-06
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引用 92 楼 GKatHere 的回复:
[quote=引用 86 楼 赵4老师 的回复:]使用#78楼的数据，在我#41楼代码的基础上优化了一下，运行不到一分钟： 
//二维数组中找最大特征数组
//http://bbs.csdn.net/topics/399169928
#pragma warning(disable:4996) //开头加这句或项目、属性、配置属性、C/C++、预处理器、预处理器定义中添加“_CRT_SECURE_NO_WARNINGS”
#include <conio.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <search.h>
#include <time.h>

#define N 50 //5000
static char a[N][N]={
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
};
int v[8],u[8],w[N],W,bestv[8],y;
int maxfitnum=0;
char vstr[1024*1024*32][4*8];//1GB
int V=0,CNTold=0;
int compare(const void *arg1,const void *arg2) {
   return strncmp((char *)arg1,(char *)arg2,4*8);
}
bool isOldv(char *str) {
    void *p;
    if (V>0 && (p=bsearch((const void *)str,(const void *)&vstr[0],V,4*8,compare))!=NULL) {
        CNTold++;
        if (CNTold%100000==0) cprintf("y:%02d/%d CNTold:%d V:%d old pattern %d,%s\r",y,N,CNTold,V,((int)p-(int)&vstr[0][0])/(4*8),str);
        return true;
    } else {
        if (V>=1024*1024*32) {
//          printf("Warnning: Too many pattern!\n");
            return false;
        } else {
            strncpy(vstr[V],str,4*8);
            V++;
            if (V%100000==0) cprintf("y:%02d/%d CNTold:%d V:%d new pattern:%s\r",y,N,CNTold,V,str);
            return false;
        }
    }
}
unsigned long ulrand(void) {
    return (
     (((unsigned long)rand()<<24)&0xFF000000ul)
    |(((unsigned long)rand()<<12)&0x00FFF000ul)
    |(((unsigned long)rand()    )&0x00000FFFul));
}

int first1(int i) {
    int j;
    for (j=0;j<N;j++) {
        if (a[i][j]==1) break;
    }
    return j;
}

void next1(int i,int &p) {
    int j;

    if (p>=N) {p=N; return;}
    for (j=p+1;j<N;j++) {
        if (a[i][j]==1) break;
    }
    p=j;
}

void test() {
    int fitnum=0,b;
    for (int i=0;i<N;i++) {
        for (int j=first1(i);j<N-v[7];next1(i,j)) {
            b=a[i][v[0]+j]+a[i][v[1]+j]+a[i][v[2]+j]+a[i][v[3]+j]+a[i][v[4]+j]+a[i][v[5]+j]+a[i][v[6]+j]+a[i][v[7]+j];
            if (b==8) fitnum++;
        }
    }
    if (fitnum>maxfitnum) {
        bestv[0]=v[0];bestv[1]=v[1];bestv[2]=v[2];bestv[3]=v[3];bestv[4]=v[4];bestv[5]=v[5];bestv[6]=v[6];bestv[7]=v[7];
        maxfitnum=fitnum;
        printf("\nfound a new bestv,maxfitnum:%d\n",maxfitnum);
//  } else {
//      printf("fitnum:%d\n",fitnum);
    }
}

void recursion(int m,int L) {
    if (L==8) {
/*
        printf("y:%04d u[0..7]:",y);
        for (int i=0;i<8;i++) {
            printf("%5d",u[i]);
        }
//      printf("\n");
*/

        for (int i=1;i<8;i++) v[i]=u[i]-u[0];
        v[0]=0;
/*
        printf("|y:%04d v[0..7]:",y);
        for (int i=0;i<8;i++) {
            printf("%5d",v[i]);
        }
        printf("\n");
*/
        char ln[4*8+1];
        for (int i=0;i<8;i++) sprintf(ln+i*4,"%04d",v[i]);
        if (!isOldv(ln)) test();
        return;
    }
    for (int i=m;i<=W-8+L;i++) {
        u[L]=w[i];
        recursion(i+1,L+1);
    }
}

void findpattern(int y) {
    W=0;
    for (int j=0;j<N;j++) {
        if (a[y][j]==1) {
            w[W]=j;
            W++;
        }
    }
    if (W>=8) {
/*      printf("y:%04d W:%5d:",y,W);
        for (int j=0;j<W;j++) {
            printf("%5d",w[j]);
        }
        printf("\n");
*/      recursion(0,0);
    }
}

int main() {
//  srand(time(NULL));
    srand(0);
/*
    for (int i=0;i<N;i++) {
        for (int j=0;j<N;j++) {
            if ((ulrand()%N) < 3) a[i][j]=1;
//          printf("%d",a[i][j]);
        }
//      printf("\n");
    }
*/
    for (y=0;y<N;y++) {
        findpattern(y);
    }

    if (maxfitnum==0) {
        printf("\nCan not find pattern!");
    } else {
        printf("\nbestv:");
        for (int i=0;i<8;i++) {
            printf("%6d ",bestv[i]);
        }
    }
    return 0;
}
//
//found a new bestv,maxfitnum:450
//y:48/50 CNTold:26600000 V:346104 old pattern 182573,00000006000800140016001800340036
//bestv:     0      2      4      6      8     10     12     14
[/quote] b[8]中的值不可重复
日立奔腾浪潮微软松下联想 2021-05-06
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可以考虑把每行看作一个二进制数字，对按行排序过的矩阵再做处理？这样可能不需要C(5000, 8)次循环了，只需要循环5000次，每次处理5000、4999... 1行数据
GKatHere 2021-05-05
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引用 86 楼 赵4老师 的回复:
使用#78楼的数据，在我#41楼代码的基础上优化了一下，运行不到一分钟： 
//二维数组中找最大特征数组
//http://bbs.csdn.net/topics/399169928
#pragma warning(disable:4996) //开头加这句或项目、属性、配置属性、C/C++、预处理器、预处理器定义中添加“_CRT_SECURE_NO_WARNINGS”
#include <conio.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <search.h>
#include <time.h>

#define N 50 //5000
static char a[N][N]={
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
};
int v[8],u[8],w[N],W,bestv[8],y;
int maxfitnum=0;
char vstr[1024*1024*32][4*8];//1GB
int V=0,CNTold=0;
int compare(const void *arg1,const void *arg2) {
   return strncmp((char *)arg1,(char *)arg2,4*8);
}
bool isOldv(char *str) {
    void *p;
    if (V>0 && (p=bsearch((const void *)str,(const void *)&vstr[0],V,4*8,compare))!=NULL) {
        CNTold++;
        if (CNTold%100000==0) cprintf("y:%02d/%d CNTold:%d V:%d old pattern %d,%s\r",y,N,CNTold,V,((int)p-(int)&vstr[0][0])/(4*8),str);
        return true;
    } else {
        if (V>=1024*1024*32) {
//          printf("Warnning: Too many pattern!\n");
            return false;
        } else {
            strncpy(vstr[V],str,4*8);
            V++;
            if (V%100000==0) cprintf("y:%02d/%d CNTold:%d V:%d new pattern:%s\r",y,N,CNTold,V,str);
            return false;
        }
    }
}
unsigned long ulrand(void) {
    return (
     (((unsigned long)rand()<<24)&0xFF000000ul)
    |(((unsigned long)rand()<<12)&0x00FFF000ul)
    |(((unsigned long)rand()    )&0x00000FFFul));
}

int first1(int i) {
    int j;
    for (j=0;j<N;j++) {
        if (a[i][j]==1) break;
    }
    return j;
}

void next1(int i,int &p) {
    int j;

    if (p>=N) {p=N; return;}
    for (j=p+1;j<N;j++) {
        if (a[i][j]==1) break;
    }
    p=j;
}

void test() {
    int fitnum=0,b;
    for (int i=0;i<N;i++) {
        for (int j=first1(i);j<N-v[7];next1(i,j)) {
            b=a[i][v[0]+j]+a[i][v[1]+j]+a[i][v[2]+j]+a[i][v[3]+j]+a[i][v[4]+j]+a[i][v[5]+j]+a[i][v[6]+j]+a[i][v[7]+j];
            if (b==8) fitnum++;
        }
    }
    if (fitnum>maxfitnum) {
        bestv[0]=v[0];bestv[1]=v[1];bestv[2]=v[2];bestv[3]=v[3];bestv[4]=v[4];bestv[5]=v[5];bestv[6]=v[6];bestv[7]=v[7];
        maxfitnum=fitnum;
        printf("\nfound a new bestv,maxfitnum:%d\n",maxfitnum);
//  } else {
//      printf("fitnum:%d\n",fitnum);
    }
}

void recursion(int m,int L) {
    if (L==8) {
/*
        printf("y:%04d u[0..7]:",y);
        for (int i=0;i<8;i++) {
            printf("%5d",u[i]);
        }
//      printf("\n");
*/

        for (int i=1;i<8;i++) v[i]=u[i]-u[0];
        v[0]=0;
/*
        printf("|y:%04d v[0..7]:",y);
        for (int i=0;i<8;i++) {
            printf("%5d",v[i]);
        }
        printf("\n");
*/
        char ln[4*8+1];
        for (int i=0;i<8;i++) sprintf(ln+i*4,"%04d",v[i]);
        if (!isOldv(ln)) test();
        return;
    }
    for (int i=m;i<=W-8+L;i++) {
        u[L]=w[i];
        recursion(i+1,L+1);
    }
}

void findpattern(int y) {
    W=0;
    for (int j=0;j<N;j++) {
        if (a[y][j]==1) {
            w[W]=j;
            W++;
        }
    }
    if (W>=8) {
/*      printf("y:%04d W:%5d:",y,W);
        for (int j=0;j<W;j++) {
            printf("%5d",w[j]);
        }
        printf("\n");
*/      recursion(0,0);
    }
}

int main() {
//  srand(time(NULL));
    srand(0);
/*
    for (int i=0;i<N;i++) {
        for (int j=0;j<N;j++) {
            if ((ulrand()%N) < 3) a[i][j]=1;
//          printf("%d",a[i][j]);
        }
//      printf("\n");
    }
*/
    for (y=0;y<N;y++) {
        findpattern(y);
    }

    if (maxfitnum==0) {
        printf("\nCan not find pattern!");
    } else {
        printf("\nbestv:");
        for (int i=0;i<8;i++) {
            printf("%6d ",bestv[i]);
        }
    }
    return 0;
}
//
//found a new bestv,maxfitnum:450
//y:48/50 CNTold:26600000 V:346104 old pattern 182573,00000006000800140016001800340036
//bestv:     0      2      4      6      8     10     12     14
源代码大师 2021-05-03
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C和C++完整教程:https://blog.csdn.net/it_xiangqiang/category_10581430.html C和C++算法完整教程:https://blog.csdn.net/it_xiangqiang/category_10768339.html
赵4老师 2021-04-30
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使用#78楼的数据，在我#41楼代码的基础上优化了一下，运行不到一分钟： 
//二维数组中找最大特征数组
//http://bbs.csdn.net/topics/399169928
#pragma warning(disable:4996) //开头加这句或项目、属性、配置属性、C/C++、预处理器、预处理器定义中添加“_CRT_SECURE_NO_WARNINGS”
#include <conio.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <search.h>
#include <time.h>

#define N 50 //5000
static char a[N][N]={
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
  {1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,0},
  {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1},
};
int v[8],u[8],w[N],W,bestv[8],y;
int maxfitnum=0;
char vstr[1024*1024*32][4*8];//1GB
int V=0,CNTold=0;
int compare(const void *arg1,const void *arg2) {
   return strncmp((char *)arg1,(char *)arg2,4*8);
}
bool isOldv(char *str) {
    void *p;
    if (V>0 && (p=bsearch((const void *)str,(const void *)&vstr[0],V,4*8,compare))!=NULL) {
        CNTold++;
        if (CNTold%100000==0) cprintf("y:%02d/%d CNTold:%d V:%d old pattern %d,%s\r",y,N,CNTold,V,((int)p-(int)&vstr[0][0])/(4*8),str);
        return true;
    } else {
        if (V>=1024*1024*32) {
//          printf("Warnning: Too many pattern!\n");
            return false;
        } else {
            strncpy(vstr[V],str,4*8);
            V++;
            if (V%100000==0) cprintf("y:%02d/%d CNTold:%d V:%d new pattern:%s\r",y,N,CNTold,V,str);
            return false;
        }
    }
}
unsigned long ulrand(void) {
    return (
     (((unsigned long)rand()<<24)&0xFF000000ul)
    |(((unsigned long)rand()<<12)&0x00FFF000ul)
    |(((unsigned long)rand()    )&0x00000FFFul));
}

int first1(int i) {
    int j;
    for (j=0;j<N;j++) {
        if (a[i][j]==1) break;
    }
    return j;
}

void next1(int i,int &p) {
    int j;

    if (p>=N) {p=N; return;}
    for (j=p+1;j<N;j++) {
        if (a[i][j]==1) break;
    }
    p=j;
}

void test() {
    int fitnum=0,b;
    for (int i=0;i<N;i++) {
        for (int j=first1(i);j<N-v[7];next1(i,j)) {
            b=a[i][v[0]+j]+a[i][v[1]+j]+a[i][v[2]+j]+a[i][v[3]+j]+a[i][v[4]+j]+a[i][v[5]+j]+a[i][v[6]+j]+a[i][v[7]+j];
            if (b==8) fitnum++;
        }
    }
    if (fitnum>maxfitnum) {
        bestv[0]=v[0];bestv[1]=v[1];bestv[2]=v[2];bestv[3]=v[3];bestv[4]=v[4];bestv[5]=v[5];bestv[6]=v[6];bestv[7]=v[7];
        maxfitnum=fitnum;
        printf("\nfound a new bestv,maxfitnum:%d\n",maxfitnum);
//  } else {
//      printf("fitnum:%d\n",fitnum);
    }
}

void recursion(int m,int L) {
    if (L==8) {
/*
        printf("y:%04d u[0..7]:",y);
        for (int i=0;i<8;i++) {
            printf("%5d",u[i]);
        }
//      printf("\n");
*/

        for (int i=1;i<8;i++) v[i]=u[i]-u[0];
        v[0]=0;
/*
        printf("|y:%04d v[0..7]:",y);
        for (int i=0;i<8;i++) {
            printf("%5d",v[i]);
        }
        printf("\n");
*/
        char ln[4*8+1];
        for (int i=0;i<8;i++) sprintf(ln+i*4,"%04d",v[i]);
        if (!isOldv(ln)) test();
        return;
    }
    for (int i=m;i<=W-8+L;i++) {
        u[L]=w[i];
        recursion(i+1,L+1);
    }
}

void findpattern(int y) {
    W=0;
    for (int j=0;j<N;j++) {
        if (a[y][j]==1) {
            w[W]=j;
            W++;
        }
    }
    if (W>=8) {
/*      printf("y:%04d W:%5d:",y,W);
        for (int j=0;j<W;j++) {
            printf("%5d",w[j]);
        }
        printf("\n");
*/      recursion(0,0);
    }
}

int main() {
//  srand(time(NULL));
    srand(0);
/*
    for (int i=0;i<N;i++) {
        for (int j=0;j<N;j++) {
            if ((ulrand()%N) < 3) a[i][j]=1;
//          printf("%d",a[i][j]);
        }
//      printf("\n");
    }
*/
    for (y=0;y<N;y++) {
        findpattern(y);
    }

    if (maxfitnum==0) {
        printf("\nCan not find pattern!");
    } else {
        printf("\nbestv:");
        for (int i=0;i<8;i++) {
            printf("%6d ",bestv[i]);
        }
    }
    return 0;
}
//
//found a new bestv,maxfitnum:450
//y:48/50 CNTold:26600000 V:346104 old pattern 182573,00000006000800140016001800340036
//bestv:     0      2      4      6      8     10     12     14
atlandisebr 2021-04-30
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是离散的8个1
跑马溜溜的山上 2021-04-30
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引用 82 楼 tong386的回复:
外层循环5000次，内层每次记录下最后一个不符合要求的位置，以及当前对比到的最后位置，移动后只需要对比新增的位点，单纬度的特征提取。这样应该效率很高。你的题目好像是找连续的8个1啊
似乎可以是离散的八个1。实际上是任意抽8列，形成5000字节，里面的ff最多
atlandisebr 2021-04-30
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可以付费开发
atlandisebr 2021-04-30
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期待优化的算法，能在30天内算完就OK！
赵4老师 2021-04-30
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引用 87 楼 早打大打打核战争 的回复:
老赵的新代码确实不会死循环了 但是性能不行啊，在我的电脑上运行时间32秒。50x50数组和5000x5000数组的时间复杂度可不是1:10000的关系，大约是1:10^17的关系，所以你的代码要计算5000x5000数组在PC上需要大概1000亿年~
这就是穷举法的“绝对优势”！
日立奔腾浪潮微软松下联想 2021-04-30
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老赵的新代码确实不会死循环了
但是性能不行啊，在我的电脑上运行时间32秒。50x50数组和5000x5000数组的时间复杂度可不是1:10000的关系，大约是1:10^17的关系，所以你的代码要计算5000x5000数组在PC上需要大概1000亿年~
赵4老师 2021-04-29
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审题！ 人家找的是同一行内8个点的分布特征，不是你画的二维特征。
tong386 2021-04-29
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我想了个好方法，你把你的a[5000][5000]，画一个5000*5000方格的图片，1用红色填充，0用绿色填充，然后把图片导出来，用人眼看，找出可能的特征，然后再用程序验证。
tong386 2021-04-29
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一张更好的图片，图片很小，你的a[5000][5000]那么大，查找量太大了。 如图：
tong386 2021-04-29
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企图通过遍历查找特征是不可能的啊。 关键点在于你的b[8]是不确定，初期最好只检查b[8]的关键几个点的值。，然后你就可以进行移动扫描了。 让我们来举一个具体的例子，找出下面图中的最多的特征： 那么按照你的遍历算法，能找出来吗？
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青少年C++编程NOIP/CSP竞赛通关语法-二维数组
1、本课程为零基础C++语法专题课，是青少儿参加信息学奥赛NOIP/CSP需要学习的语法专题课； 2、本课程每节课均配有练习题以及真题，以锻炼孩子的解题能力；... 6、本课程介绍的是二维数组。也是历来竞赛重要考点。
C语言总结（一维数组、二维数组、字符数组和字符串）
C语言总结第七章、数组一维数组一维数组的定义一维数组的引用一维数组的初始化程序举例二维数组及多维数组二维数组的定义二维数组元素的引用二维数组元素的初始化程序举例字符数组和字符串字符数组 第七章、数组 ...
java二维数组中找最大值和最小值_求二维数组a中的最大值和最小值
展开全部//参考代码如下：#includeint main(){int i,j,max,min,a[2][3] = {{1,2,3},{9,8,7}};max=min=a[0][0];...i++)//依次判断二维数组中的每e69da5e887aa62616964757a686964616f31333431353964一个数for(...
Java 学习 二维数组练习 定义一个5X5的二维数组。 然后使用随机数填充该二维数组。找出这个二维数组里，最大的那个值，并打印出其二维坐标
找出这个二维数组里，最大的那个值，并打印出其二维坐标 import java.util.Arrays; import java.util.Random; public class Demo06 { /* 定义一个5X5的二维数组。 然后使用随机数填充该二维数组。 找出这个二维...
PTA-求二维数组中的最大值和最小值
求二维数组中的最大值和...输入9个整型数到二维数组a[3][3]，数据之间用一个空格分隔。 输出格式: 输出数组a的最大值和最小值 输入样例: 在这里给出一组输入。例如： 1 2 3 4 5 6 7 8 9 输出样例: 在这里给...
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