如何计算数列和的绝对值最小

[fengnetman]

现有一个数组，里面包含了正数和负数，取其中n个数，要求这n个数的和的绝对值是最小的！

[LeonTown]

[Quote=引用 106 楼 linren 的回复:]
（上接103楼）
【一个新的测试用例】
int g_iArr[] ={-96,31,32,33,1,-2,5};
GetSubArr(4, num);
Min Abs Sum: 35
They are:31 1 -2 5
Press any key to continue

答案应该是：
[33, 32, 31, -96], |SUM|=0
[/Quote]

如果在我的基础上，
先对数列按绝对值，做一下从小到大的排序，
会有效果吗？

[linren]

[Quote=引用 104 楼 LeonTown 的回复:]
嗯，是有问题。
谢谢！
[/Quote]
这个算法在求近似解上的实力真的很强
如果两者能整合在一起就好了……

[Quote=引用 105 楼 fengnetman 的回复:]
确实，向前取最小的时候是有问题，疏忽了。看来还得继续讨论
[/Quote]
感觉上有可能的确是NP难问题……

[linren]

[Quote=引用 107 楼 LeonTown 的回复:]
如果在我的基础上，
先对数列按绝对值，做一下从小到大的排序，
会有效果吗？
[/Quote]
恩
的确是有效果！

[LeonTown]

[Quote=引用 109 楼 linren 的回复:]
引用 107 楼 LeonTown 的回复:
如果在我的基础上，
先对数列按绝对值，做一下从小到大的排序，
会有效果吗？


恩
的确是有效果！
[/Quote]

也只是对上述两个错误的例子。
不知道，还有没有其它错误的例子。。。

[linren]

[Quote=引用 107 楼 LeonTown 的回复:]
如果在我的基础上，
先对数列按绝对值，做一下从小到大的排序，
会有效果吗？
[/Quote]
还是有一些问题……
int g_iArr[] ={1,-2,5,31,32,33,-96};
GetSubArr(3, num);
Min Abs Sum: 31
They are:32 33 -96
Press any key to continue

答案是：
n=3:
[1, -2, 5], |SUM|=4
Press any key to continue

[LeonTown]

[Quote=引用 111 楼 linren 的回复:]
引用 107 楼 LeonTown 的回复:
如果在我的基础上，
先对数列按绝对值，做一下从小到大的排序，
会有效果吗？


还是有一些问题……
int g_iArr[] ={1,-2,5,31,32,33,-96};
GetSubArr(3, num);
Min Abs Sum: 31
They are:32 33 -96
Press any key to continue

答案是：
n=3:
[1, -2, 5], |SUM|=4
Press any key to continue
[/Quote]

那针对这个case，
把遍历顺序，改为从前向后，
呵呵。。。

[LeonTown]

可能这道题，
不能简单的使用传统方法

[LeonTown]

而且，上述的错误例子，
都是答案集中在数列的两端

[linren]

[Quote=引用 113 楼 LeonTown 的回复:]
可能这道题，
不能简单的使用传统方法
[/Quote]
恩
的确是很有可能

推荐一个很好的题：
彩票N保M缩水问题（最小集合覆盖问题）
这个问题虽然没有经过推荐不过的确是很不错

[tanwan]

我41楼回答的大家看看额...
1,-2,5,31,32,33,-96 这组数据我说的好象能过~~~

[fengnetman]

你的方法类似LeonTown的，看100楼的代码。这个已经发现有问题了。见103楼
现在提供一个新的想法：首先去着手求各个数之间的差值（两两之间），再根据差值来获得结果

[linren]

【思路】
利用求近似解的思路优化枚举数组，这样比单纯的按绝对值排序得到的效果要好上很多

【程序】

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <windows.h>
/***只用替换这里*****************/
#define ASIZE 200 
int a[ASIZE]={-176, 412, 437, -36, 164, -33, -393, 285,
-279, -290, 217, -325, -316, 51, -377, -195,
318, -141, -277, -18, 208, 364, 364, -453,
188, -2, -45, -343, 115, -309, -303, 240,
147, -331, 318, 383, -324, 272, -253, -36,
-419, -173, -392, -213, -55, 119, -50, 1,
-96, 467, 127, 106, -221, -40, 222, 300,
143, -125, -76, 468, -13, -199, 269, -417,
-361, -487, 152, -195, -409, 119, 5, 61,
-114, -484, -438, 384, 229, -321, 106, -213,
480, -174, 113, -29, -330, -24, -312, -376,
212, -55, -287, -386, 458, 300, 103, -155,
262, 312, 80, 56, -314, 428, -132, 357,
-458, 308, -272, -233, -115, -60, -285, -454,
147, -115, 89, -397, 47, 161, -233, -249,
-60, -180, -260, 98, -232, -273, -162, 361,
255, -406, -225, 367, -481, -223, -141, -461,
-264, -106, 289, -358, 40, -231, 430, -432,
286, -253, -8, 438, -142, -80, 495, -176,
155, 267, -51, 332, -67, 275, 142, 374,
143, 39, 442, 374, 428, 398, -471, 201,
-63, 121, -45, 125, -397, -314, -64, 47,
-439, -327, 294, 316, -167, 312, -60, -112,
-35, 211, -180, 131, -245, -207, -284, 33,
-271, 486, -184, 120, -465, 179, 461, -156};
/********************************/

int compar(const void *a,const void *b){
	int *x=(int*)a;
	int *y=(int*)b;
	if(*x>*y) return 1;
	if(*x==*y) return 0;
	return -1;
}

struct level{
	int *now;
	int len;
};

void add(struct level *llv,int n){
	int i=0;
	llv->len=n;
	llv->now=(int*)malloc(sizeof(int)*n);
	memset((char*)(llv->now),0,sizeof(int)*n);
	for(i=0;i<n;i++){
		llv->now[i]=i;
	}
}

void del(struct level *llv){
	free(llv->now);
}

int timeout;
void mytimer(LPVOID sc){
	int seconds;
	seconds=(*(int*)sc)*1000;
	Sleep(seconds);
	timeout=0;
}

void fun(int *array,int size,int n,unsigned int seconds){
	HANDLE thrd;
	int i,j,k;
	int x,y,u,v,o;
	int z,z2;
	int *a;
	int *b;
	int *c;
	char *mask;
	struct level p;
	if(size<=0||size<n||n<=0){
		printf("size<=0||size<n||n<=0 error!\n");return;
	}

	a=(int*)malloc(sizeof(int)*size);
	b=(int*)malloc(sizeof(int)*n);
	c=(int*)malloc(sizeof(int)*size);
	mask=(char*)malloc(sizeof(char)*size);
	memcpy(a,array,sizeof(int)*size);
	memset(b,0,sizeof(int)*n);
	memset(c,0,sizeof(int)*size);
	memset((char*)&p,0,sizeof(struct level));

	qsort(a,size,sizeof(int),compar);
	if(a[0]>=0){
		printf("array error: array need <0 number\n");
		free(mask);free(c);free(b);free(a);return;
	}
	if(a[size-1]<=0){
		printf("array error: array need >0 number\n");
		free(mask);free(c);free(b);free(a);return;
	}

	timeout=1;
	thrd=CreateThread(NULL,0,(LPTHREAD_START_ROUTINE)mytimer,(LPVOID)&seconds,0,NULL);

	if(n==1){
		for(i=0;i<size;i++){
			if(i==0){
				z=abs(a[i]);b[0]=a[0];continue;
			}
			z2=abs(a[i]);
			if(z>z2){
				z=z2;b[0]=a[i];
			}
		}
	}
	else if(n==size){
		z2=0;
		for(i=0;i<n;i++){
			b[i]=a[i];z2+=a[i];
		}
		z=abs(z2);
	}
	else if(n<=size/2){
		/**找一个近似解并且优化枚举数组***********************************/
		o=1;
		for(i=0;i<size;i++){
			if(a[i]>=0){
				y=i;break;
			}
		}
		x=y-1;z2=0;
		for(i=0;i<size;i++){
			if(x>=0&&y<size){
				u=a[x]+z2;
				v=a[y]+z2;
				if(abs(u)<abs(v)){
					z2=u;
					if(i==n-1){
						z=abs(z2);b[i]=a[x];
					}
					else if(i<n-1) b[i]=a[x];
					c[i]=a[x];x--;
					continue;
				}else{
					z2=v;
					if(i==n-1){
						z=abs(z2);b[i]=a[y];
					}
					else if(i<n-1) b[i]=a[y];
					c[i]=a[y];y++;
					continue;
				}
			}
			if(x>=0){
				z2=a[x]+z2;
				if(i==n-1){
					b[i]=a[x];z=abs(z2);
				}
				else if(i<n-1) b[i]=a[x];
				c[i]=a[x];x--;
				continue;
			}else{
				z2=a[y]+z2;
				if(i==n-1){
					b[i]=a[y];z=abs(z2);
				}
				else if(i<n-1) b[i]=a[y];
				c[i]=a[y];y++;
				continue;
			}
		}
		/***********************************找一个近似解并且优化枚举数组**/
		if(z>0){
		/**选择****************/
			add(&p,n);
			while(timeout){
				z2=0;
				for(i=0;i<n;i++){
					z2+=c[p.now[i]];
				}
				z2=abs(z2);
				if(z>z2){
					z=z2;
					for(i=0;i<n;i++) b[i]=c[p.now[i]];
					if(z==0) break;
				}
				if(p.now[p.len-1]==size-1){
					for(i=1,j=p.len-2;j>=0;j--,i++){
						if((p.now[j])<(size-1-i)) break;
					}
					if(j<0) break;
					p.now[j]++;
					k=p.now[j]+1;
					for(i=j+1;i<p.len;i++) p.now[i]=k++;
				}else{
					p.now[p.len-1]++;
				}
			}
			/****************选择**/
		}
	}
	else{
		/**找一个近似解并且优化枚举数组***********************************/
		o=1;
		for(i=0;i<size;i++){
			if(a[i]>=0){
				y=i;break;
			}
		}
		x=y-1;z2=0;
		for(i=0;i<size;i++){
			if(x>=0&&y<size){
				u=a[x]+z2;
				v=a[y]+z2;
				if(abs(u)<abs(v)){
					z2=u;
					if(i==n-1){
						z=abs(z2);b[i]=a[x];
					}
					else if(i<n-1) b[i]=a[x];
					c[i]=a[x];x--;
					continue;
				}else{
					z2=v;
					if(i==n-1){
						z=abs(z2);b[i]=a[y];
					}
					else if(i<n-1) b[i]=a[y];
					c[i]=a[y];y++;
					continue;
				}
			}
			if(x>=0){
				z2=a[x]+z2;
				if(i==n-1){
					b[i]=a[x];z=abs(z2);
				}
				else if(i<n-1) b[i]=a[x];
				c[i]=a[x];x--;
				continue;
			}else{
				z2=a[y]+z2;
				if(i==n-1){
					b[i]=a[y];z=abs(z2);
				}
				else if(i<n-1) b[i]=a[y];
				c[i]=a[y];y++;
				continue;
			}
		}
		memset(a,0,sizeof(int)*size);
		for(i=0;i<size;i++) a[i]=c[size-i-1];
		/***********************************找一个近似解并且优化枚举数组**/
		if(z>0){
		/**排除****************/
			add(&p,size-n);
			while(timeout){
				memset(mask,0,sizeof(char)*size);
				for(i=0;i<size-n;i++) mask[p.now[i]]=1;
				z2=0;
				for(i=0;i<size;i++) if(mask[i]==0) z2+=a[i];
				z2=abs(z2);
				if(z>z2){
					z=z2;j=0;
					for(i=0;i<size;i++) if(mask[i]==0) b[j++]=a[i];
					if(z==0) break;
				}
				if(p.now[p.len-1]==size-1){
					for(i=1,j=p.len-2;j>=0;j--,i++){
						if((p.now[j])<(size-1-i)) break;
					}
					if(j<0) break;
					p.now[j]++;
					k=p.now[j]+1;
					for(i=j+1;i<p.len;i++) p.now[i]=k++;
				}else{
					p.now[p.len-1]++;
				}
			}
			/**排除****************/
		}
	}

	printf("n=%d:\n",n);
	if(timeout) printf("[");
	else printf("time out! the best test answer find in %d seconds is:\n[",seconds);
	for(i=0;i<n-1;i++){
		printf("%d, ",b[i]);
	}
	printf("%d], |SUM|=%d\n",b[i],z);
	TerminateThread(thrd,NULL);
	del(&p);free(mask);free(c);free(b);free(a);
}

void main(){
	int i;
	for(i=1;i<=ASIZE;i++) fun(a,ASIZE,i,10);
}

【运行结果】
大部分情况都找到最优解了，超时的只有以下几种情况:
n=190: |SUM|=177
n=191: |SUM|=616
n=192: |SUM|=1069
n=193: |SUM|=1523
n=194: |SUM|=1981
n=195: |SUM|=2442
n=196: |SUM|=2907
这些超时的答案和LeonTown所求的答案是一样的

[linren]

（上接118楼）

【测试数据】
-176, 412, 437, -36, 164, -33, -393, 285,
-279, -290, 217, -325, -316, 51, -377, -195,
318, -141, -277, -18, 208, 364, 364, -453,
188, -2, -45, -343, 115, -309, -303, 240,
147, -331, 318, 383, -324, 272, -253, -36,
-419, -173, -392, -213, -55, 119, -50, 1,
-96, 467, 127, 106, -221, -40, 222, 300,
143, -125, -76, 468, -13, -199, 269, -417,
-361, -487, 152, -195, -409, 119, 5, 61,
-114, -484, -438, 384, 229, -321, 106, -213,
480, -174, 113, -29, -330, -24, -312, -376,
212, -55, -287, -386, 458, 300, 103, -155,
262, 312, 80, 56, -314, 428, -132, 357,
-458, 308, -272, -233, -115, -60, -285, -454,
147, -115, 89, -397, 47, 161, -233, -249,
-60, -180, -260, 98, -232, -273, -162, 361,
255, -406, -225, 367, -481, -223, -141, -461,
-264, -106, 289, -358, 40, -231, 430, -432,
286, -253, -8, 438, -142, -80, 495, -176,
155, 267, -51, 332, -67, 275, 142, 374,
143, 39, 442, 374, 428, 398, -471, 201,
-63, 121, -45, 125, -397, -314, -64, 47,
-439, -327, 294, 316, -167, 312, -60, -112,
-35, 211, -180, 131, -245, -207, -284, 33,
-271, 486, -184, 120, -465, 179, 461, -156

【测试结果】
n=1: |SUM|=1
n=2~189: |SUM|=0
n=190: |SUM|=177 （timeout）
n=191: |SUM|=616 （timeout）
n=192: |SUM|=1069 （timeout）
n=193: |SUM|=1523 （timeout）
n=194: |SUM|=1981 （timeout）
n=195: |SUM|=2442 （timeout）
n=196: |SUM|=2907 （timeout）
n=197: |SUM|=3378
n=198: |SUM|=3859
n=199: |SUM|=4343
n=200: |SUM|=4830

[java1109]

mark

[tanwan]

[Quote=引用 103 楼 linren 的回复:]
引用 100 楼 LeonTown 的回复:
C/C++ code
#include <vector>
using namespace std;

int g_iArr[] = //{-5,-2,-1,3,5}; //原始序列
//{-5, 4, 9, 3, 6, 1};
{
-176, 412, 437, -36, 164, -33, -393, 285,
-279, -290, 217, -325, -316, 51, -377, -195,
318, -141, -277, -18, 208, 364, 364, -453,
188, -2, -45, -343, 115, -309, -303, 240,
147, -331, 318, 383, -324, 272, -253, -36,
-419, -173…
[/Quote]
这组测试案例的最小值是１？如下
【测试用例】
int g_iArr[] = {-100,49,51,-1,2,4,-5};
GetSubArr(3, num);

【测试结果】
Min Abs Sum: 1
They are:2 4 -5
Press any key to continue
应该是0吧？-100+49+51=0

[linren]

刚才用10989测试了一下，发现这个数在32位（10进制）下是没有解的。测试需要的时间约为1个小时。

【程序】

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <windows.h>

void int_to_big(char* a,int x,int len){
	int i,c;
	c=0;
	memset(a,0,sizeof(char)*len);
	for(i=0;i<len;i++){
		if(x<10){
			a[i]=x;return;
		}
		c=x%10;a[i]=c;x/=10;
	}
}

void print_big(char* a){
	int i,c;
	c=0;
	for(i=31;i>=0;i--){
		if(a[i]!=0){
			if(c==0) c=1;
			printf("%d",a[i]);
		}else{
			if(c!=0) printf("0");
		}
	}
	if(c==0) printf("0");
	printf("\n");
}

void mask_to_msk(int mask,char* msk){
	int i;
	memset(msk,0,sizeof(char)*32);
	for(i=0;i<32;i++){
		if(mask&(1<<i)) msk[i]=1;
	}
}
void msk_to_mask(int* mask,char* msk){
	int i;
	*mask=0;
	for(i=0;i<32;i++) if(msk[i]==1) *mask|=(1<<i);
}

int init_mask(char* a,int* mask,char* msk){
	int i,c;
	char in[32];
	memset(in,0,sizeof(in));
	memset(msk,0,sizeof(char)*32);c=-1;
	for(i=31;i>=0;i--){
		if(a[i]!=0&&c==-1) c=i;
		if(a[i]>1) break;
	}
	if(i==-1) return 1;
	if(c+1==32) return -1;
	msk[c+1]=1;
	msk_to_mask(mask,msk);
	return 0;
}

int is_answer(char* msk,int xx){
	int i,j,lm;
	int cc;
	int b[8],d;
	memset(b,0,sizeof(b));
	for(i=31;i>=0;i--) if(msk[i]!=0) break;
	lm=i+1;
	if(lm%4==0) cc=lm/4;
	else cc=lm/4+1;
	for(i=0;i<cc;i++){
		for(j=0;j<4;j++){
			b[i]=b[i]*10+msk[i*4+3-j];
		}
	}
	d=b[cc-1]%xx;
	for(i=cc-2;i>=0;i--){
		d=d*10000+b[i];
		d=d%xx;
	}
	if(d==0){
		return 1;
	}
	return 0;
}

unsigned prt;
void fun(int x){
	char *pa,*msk;
	int mask,k;
	pa=(char*)malloc(sizeof(char)*32);
	msk=(char*)malloc(sizeof(char)*32);

	//把x转换成10进制表示，并存入pa中
	memset(pa,0,sizeof(char)*32);
	int_to_big(pa,x,32);

	//初始化msk和mask
	k=init_mask(pa,&mask,msk);
	if(k==1){
		printf("%d=>%d\n",x,x);
		free(pa);free(msk);return;
	}else if(k==-1){
		printf("error!\n");
		free(pa);free(msk);return;
	}
	//找答案
	while(1){
		if(is_answer(msk,x)){
			printf("%d=>",x);
			print_big(msk);
			free(pa);free(msk);return;
		}
		mask++;prt=mask;
		if(mask==-1){
			printf("can not find in 32 bits!\n");break;
		}
		mask_to_msk(mask,msk);
	}

	free(pa);free(msk);
}


void mytimer(LPVOID sc){
	unsigned myprt;
	unsigned total=4294967295;
	while(1){
		Sleep(30000);
		myprt=prt;
		printf("test to: %.3f%%\n",100.0*myprt/(1.0*total));
		GetTickCount();
	}
}
int main(){
	HANDLE thrd;
    time_t t1,t2;
    struct tm *pt;

	thrd=CreateThread(NULL,0,(LPTHREAD_START_ROUTINE)mytimer,NULL,0,NULL);

	time(&t1);
	pt=localtime(&t1);
	printf("start\t%02d:%02d:%02d\n",pt->tm_hour,pt->tm_min,pt->tm_sec);
	
	fun(10989);
	TerminateThread(thrd,NULL);

	time(&t2);
	pt=localtime(&t1);
	printf("start\t%02d:%02d:%02d\n",pt->tm_hour,pt->tm_min,pt->tm_sec);
	pt=localtime(&t2);
	printf("end\t%02d:%02d:%02d\n",pt->tm_hour,pt->tm_min,pt->tm_sec);
    return 0;
}

【运行结果】
start 14:25:38
test to: 0.899%
test to: 1.781%
test to: 2.665%
test to: 3.545%
test to: 4.434%
test to: 5.321%
test to: 6.204%
test to: 7.040%
test to: 7.869%
test to: 8.700%
test to: 9.526%
test to: 10.355%
test to: 11.181%
test to: 12.004%
test to: 12.830%
test to: 13.669%
test to: 14.507%
test to: 15.341%
test to: 16.178%
test to: 17.013%
test to: 17.846%
test to: 18.675%
test to: 19.508%
test to: 20.340%
test to: 21.170%
test to: 21.996%
test to: 22.827%
test to: 23.653%
test to: 24.479%
test to: 25.307%
test to: 26.149%
test to: 26.990%
test to: 27.828%
test to: 28.668%
test to: 29.509%
test to: 30.346%
test to: 31.178%
test to: 32.019%
test to: 32.857%
test to: 33.702%
test to: 34.537%
test to: 35.375%
test to: 36.209%
test to: 37.041%
test to: 37.874%
test to: 38.713%
test to: 39.549%
test to: 40.385%
test to: 41.222%
test to: 42.059%
test to: 42.893%
test to: 43.721%
test to: 44.556%
test to: 45.389%
test to: 46.222%
test to: 47.051%
test to: 47.883%
test to: 48.712%
test to: 49.535%
test to: 50.370%
test to: 51.214%
test to: 52.060%
test to: 52.904%
test to: 53.750%
test to: 54.594%
test to: 55.437%
test to: 56.275%
test to: 57.121%
test to: 57.966%
test to: 58.812%
test to: 59.652%
test to: 60.485%
test to: 61.323%
test to: 62.159%
test to: 63.000%
test to: 63.846%
test to: 64.691%
test to: 65.533%
test to: 66.376%
test to: 67.223%
test to: 68.065%
test to: 68.902%
test to: 69.743%
test to: 70.583%
test to: 71.424%
test to: 72.259%
test to: 73.095%
test to: 73.929%
test to: 74.760%
test to: 75.599%
test to: 76.442%
test to: 77.283%
test to: 78.122%
test to: 78.963%
test to: 79.803%
test to: 80.642%
test to: 81.479%
test to: 82.322%
test to: 83.161%
test to: 84.001%
test to: 84.839%
test to: 85.675%
test to: 86.509%
test to: 87.341%
test to: 88.179%
test to: 89.017%
test to: 89.855%
test to: 90.690%
test to: 91.530%
test to: 92.366%
test to: 93.197%
test to: 94.030%
test to: 94.868%
test to: 95.704%
test to: 96.538%
test to: 97.370%
test to: 98.203%
test to: 99.035%
test to: 99.869%
can not find in 32 bits!
start 14:25:38
end 15:25:13
Press any key to continue

[linren]

上面发错地方了（汗）……

[fengnetman]

今天结贴吧，不过有些话还是要说一下。就目前Csdn的算法版，个人觉得已大不如前，几位大牛基本也无法在算法版看到了。一个帖子发出来，也基本没几个人会参与谈论，这里的种种弊端我也就不说了。还有有些人发的贴子完全就是学生的作业贴，已完全失去了意义，哎，不说了，希望Csdn能再次活跃起来吧。
这里还是很感谢积极回帖的几位！

[fqhlysliuyansen]

哪有这么容易 的

[Quote=引用 7 楼 的回复:]

用Dp可以，本质还是一个01背包，按照5楼的方法，分成两个集合，效率应该会更好一些，有些分治的意味！
[/Quote]

[fqhlysliuyansen]

你太搞笑了



[Quote=引用 35 楼 的回复:]

这和求最大子数列的问题的本质是一样的，在网上找吧，一大堆的解法
[/Quote]