求解：华为一道面试题

2007-09-07 01:44:44 2.4k 阅读 AI 写同款 · GEO 优化AI 选题数据监测一键分发 ›

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面试题-华为-空瓶换饮料问题-1 面试-华为-空瓶换饮料问题-1 有这样一道智力题：“某商店规定：三个空汽水瓶可以换一瓶汽水。小张手上有十个空汽水瓶，她最多可以换多少瓶汽水喝？” 阅读详情

一个五位数字ABCDE*4=EDCBA，这五个数字不重复，请编程求出来，谢谢！希望有代码，我不知道如何下笔！

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华为面试题（小朋友高矮排序，要求移动距离最小）-java版题目描述先有一队小朋友，他们高矮不同，我们以正整数数组表示这一队小朋友的身高，如数组{5,3,1,2,3}。我们现在希望小朋友排队，以“高”“矮”“高”“矮”顺序排列，每一个“高”位置的小朋友要比相邻位置高或者相等；每一个“矮”位置的小朋友要比相邻的位置矮或者相等；要求小朋友们移动的距离和最小，第一个从“高”位开始排，输出最小移动距离即可。例如，在示范小队{5,3,1,2,3}中，{5,1,3,2,3}是排序结果。{5,2,3,1,3}虽然也满足“高”“矮”“高”“矮”顺序排列，但小朋友们的移动距离阅读详情

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切换为时间正序

afei0514 2008-04-14

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谢谢各位这么热心，开始只给了20分，没有给分的对不起。

relishou 2007-09-13

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int main()
{
int a,b,c,d,e,num;
for(int i=10001;i<33333;i++)
{
e=i%10;
d=i/10%10;
c=i/100%10;
b=i/1000%10;
a=i/10000;
if((i*4==(e*10000+d*1000+c*100+b*10+a))&(a!=b)&(b!=c)&(c!=d)&(d!=e))
{num=i;
cout<<"The number is "<<num<<"\n";
break;}
}
return 0;
}

diyemei 2007-09-13

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#include <iostream>
#include <cstdlib>

int main()
{
int A,B,C,D,E;
int i,j;
for(A=1; A<10;A++)
for(B=0;B!=A&&B<10;B++)
for(C=0;C!=B&&C!=A&&C<10;C++)
for(D=0;D!=C&&D!=B&&D!=A&&D<10;D++)
for(E=1;E!=D&&E!=C&&E!=B&&E!=A&&E<10;E++)
{
i = 4 * (10000 * A + 1000 * B + 100 * C + 10 * D + E);
j = 10000 * E + 1000 * D + 100 * C + 10 * B + A;
if(i == j)printf("%d%d%d%d%d\n",A,B,C,D,E);
}
system("pause");
return 0;
}

我写的好象得不出结果？？？
怎么回事情啊？
我用的是DEV-CPP

xkw365 2007-09-13

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这个没难度，试试那个去：http://community.csdn.net/Expert/topic/5756/5756076.xml?temp=.2979853

^_^

czk037 2007-09-12

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看得都眼花了。。。
都是牛人。。。

endofhero 2007-09-12

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留名

clxye 2007-09-12

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能推理出来，真是佩服。
我也觉得这应该是考验一个人的遇事的分析和推理能力。

yixiao386 2007-09-12

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我想到的也是这个方法
#include <iostream>
using namespace std;

int main(int argc, char* argv[])
{
int A = 2, B, C, D, E = 8;
long int temp1, temp2;

for(B = 0; B <= 9; B++)
{
if((B != A) && (B != E))
{
for(C = 0; C <= 9; C++)
{
if((C != A) && (C != B) && (C != E))
{
for(D = 0; D <= 9; D++)
{
if((D != A) && (D != B) && (D != C) && (D != E))
{
temp1 = A * 10000 + B * 1000 + C * 100 + D * 10 + E;
temp2 = E * 10000 + D * 1000 + C * 100 + B * 10 + A;
if((temp1 * 4) == temp2)
cout << temp1 << endl;
}
}
}
}
}

}
system("pause");

return 0;
}

manmanlai 2007-09-12

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part 1 + part 2 + part 3
------------------------
Output
Enter Count:5 <-----------------输入总共有多少位
Enter Multiplicator:4 <---------输入乘数是多少
X-1:1, 2,
X-2:4, 8,
X-3:2,
X-4:8,
X-Number:20008
X-5:0, 1, 2,
X-6:0, 1,
X-7:2, 7,
X-8:7,
X-Number:21078

X-Number is:21978
21978 X 4 = 87912
--------------------------
请给分，谢谢

diyemei 2007-09-12

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manmanlai 2007-09-12

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//更新組合，將以求出的數字與可能存在的數字集合進行對比，將重復的數字去除
int updataNoRepeatNumber(int Number, int Count , char *DigitList, int time)
{
char *tmpDig = (char *)malloc(strlen(DigitList) + 1024);
char *tmpOldDig = (char *)malloc(Count + 1024);
strcpy(tmpDig, DigitList);
memset(DigitList, 0, strlen(DigitList));
memset(tmpOldDig, 0, Count + 1024);
int count = 0;
for(int i = 0; i < 2 *time; i++)
{
int tmpSourInt = 0;
char tmp[2];
tmp[1] = '\0';
if(i < time)
{
tmpSourInt = Number / (int)pow(10, Count -1 - i);
Number -= tmpSourInt * (int)pow(10, Count -1 -i);
}
else
{
if(Number < 10)
{
tmpSourInt = Number;
i = 2* time;
}
else
{
tmpSourInt = Number / (int)pow(10, i - 1);
}
Number -= tmpSourInt * (int)pow(10, i - 1);
}
itoa(tmpSourInt, tmp, 10);
strcat(tmpOldDig, tmp);
}
for(int i = 0; i < (int)strlen(tmpDig); i++)
{
char tmpChar[2];
tmpChar[1] = '\0';
tmpChar[0] = tmpDig[i];
if(strstr(tmpOldDig, tmpChar) == NULL)
{
strcat(DigitList, tmpChar);
count++;
}
}
free(tmpDig);
free(tmpOldDig);
return count;
}
//解析出另一組可能性組合中能確定的數字
int parseOtherDigit(int MultNumber, int Number, char * DigitList, char *OtherDigitList, int time)
{
char *tmpDigitList = (char *)malloc(strlen(DigitList) + 1024);
memset(tmpDigitList, 0, strlen(DigitList) + 1024);
strcpy(tmpDigitList, DigitList);
memset(DigitList, 0, strlen(DigitList));
int digitNumber = Number % 10;
char tmp[2];
int tmpInt = 0;
int tmpDigit = 0;
int count = 0;
bool isRight = false;
tmp[1] = '\0';
for(int i = 0; i < (int)strlen(tmpDigitList); i++)
{
tmp[0] = tmpDigitList[i];
tmpInt = atoi(tmp);
for(int j = 0; j < 10; j++)
{
tmpDigit = (j*4 + 3)%10;
if(tmpDigit == tmpInt)
{
char tmpChar[10];
itoa(j, tmpChar, 10);
strcat(OtherDigitList, tmpChar);
isRight = true;
count ++;
}
}
if(isRight)
{
strcat(DigitList, tmp);
}
}
free(tmpDigitList);
return count;
}
//------------------part 3

manmanlai 2007-09-12

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//解析最後一位
memset(Digit, 0, 11);
if(Count % 2 != 0)//如果是奇數位
{
index = Count /2;//前一位
int tmpDigit = (Number / (int)pow(10, index-1))%10;
for(int i = 0; i < 10; i++)
{
if((i * MultNumber + (Count * tmpDigit )/10)%10 == i)
{
char tmpChar[10];
itoa(i, tmpChar, 10);
strcat(Digit, tmpChar);
}
}
}
tmp[0] = Digit[0];
Number += atoi(tmp) * (int)pow(10, Count%2 + 1);
if(!isSuccess)
{
return 0;//無解
}
return Number;
}

//解析出沒有進位的情況
//e.g. 2*4 = 8 1*4 = 4
//Not!!!! 3*4 =12 4 * 4 = 16
int parseDigit(int MultNumber, char *DigitList, bool isFirst)
{
int count = 0;
int i = 0;
if(isFirst)
{
i = 1;
}
for(i; i < 10; i++)
{
int tenDigit = (i * MultNumber)/10;
if(tenDigit == 0)
{
char tmp[10];
itoa(i, tmp, 10);
strcat(DigitList, tmp);
count ++;
}
}
return count;
}

//解析出另一組可能性
//因爲該算法是每次確定首尾2組數字，故需通過一組可能確定另一組可能
int parseOtherPossb(int MultNumber, char *otherDigitList, char *DigitList)
{
int count = 0;
for(int i = 0; i < (int)strlen(DigitList); i++)
{
char tmp[2];
tmp[1] = '\0';
tmp[0] = DigitList[i];
int Digit = atoi(tmp);
Digit *= MultNumber;
strcpy(tmp, itoa(Digit, tmp, 10));
strcat(otherDigitList, tmp);
count ++;
}
return count;
}

//解析出只有十位的集合
int parseSingleMedian(int MultNumber, char *DigitList, char *otherDigitList)
{
int count = 0;
char tmpOther[2];
char tmpDigit[2];
tmpOther[1] = '\0';
tmpDigit[1] = '\0';

char *tmpDigitList = (char *)malloc(strlen(DigitList)+1024);
memset(tmpDigitList, 0, strlen(DigitList + 1024));
strcpy(tmpDigitList, DigitList);
memset(DigitList, 0, strlen(DigitList));

char *tmpOtherList = (char *)malloc(strlen(otherDigitList)+1024);
memset(tmpOtherList, 0, strlen(otherDigitList)+1024);
strcpy(tmpOtherList, otherDigitList);
memset(otherDigitList, 0, strlen(otherDigitList));

for(int i = 0; i < (int)strlen(tmpOtherList); i++)
{
tmpOther[0] = tmpOtherList[i];
tmpDigit[0] = tmpDigitList[i];
if((atoi(tmpOther) * MultNumber)%10 == atoi(tmpDigit))
{
strcat(otherDigitList, tmpOther);
strcat(DigitList, tmpDigit);
count++;
}
}
free(tmpDigitList);
free(tmpOtherList);
return count;
}
//---------------------part 2

manmanlai 2007-09-12

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#include "stdio.h"
#include "stdlib.h"
#include "malloc.h"
#include "string.h"
#include "math.h"

int logicParse(int Count, int MultNumber);
int parseDigit(int MultNumber, char *DigitList, bool isFirst = false);
int parseOtherPossb(int MultNumber, char *otherDigitList, char *DigitList);
int parseSingleMedian(int MultNumber, char *DigitList, char *otherDigitList);
int updataNoRepeatNumber(int Number, int Count, char *DigitList, int time);
int parseOtherDigit(int MultNumber, int Number, char * DigitList, char *OtherDigitList, int time);

int main()
{
char *Number = (char *)malloc(512);
int Count = 0;
int Xnumber;
int MultNumber = 0;
int errorCode = 0;

loop:memset(Number, 0, 512);
printf("Enter Count:");
fgets(Number, 64, stdin);
Count = atoi(Number);

printf("Enter Multiplicator:");
fgets(Number, 10, stdin);
MultNumber = atoi(Number);
errorCode = 0;
if((Count == 0)||(MultNumber == 0))
{
printf("Enter error\n");
errorCode = 1;
}
if(Count > 9)
{
printf("Enter error\n");
errorCode = 1;
}
if(errorCode == 0)
{
Xnumber = logicParse(Count, MultNumber);
printf("\nX-Number is:%d\n", Xnumber);
printf("%d X 4 = %d\n", Xnumber, Xnumber*4);
}
getchar();
goto loop;
free(Number);
return errorCode;
}

//解析出數字,如果返回0 表示無解
int logicParse(int Count, int MultNumber)
{
int Number = (int)pow(10, Count -1);
int index = 0;
char tmp[2];
char Digit[11];
char otherDigit[11];
char endDigitList[11];
bool isSuccess = true;
memset(otherDigit, 0, 11);
memset(Digit, 0, 11);
memset(endDigitList, 0, 11);
tmp[1] = '\0';
if((MultNumber == 1)||(MultNumber == 0))
{
return 0;//無解
}
if(parseDigit(MultNumber, Digit, true) == 0)//parse the first number
{
return 0;//無解
}
printf("X-1:");
for(int i = 0; i < (int)strlen(Digit); i++)
{
printf("%c, ",Digit[i]);
}
if(parseOtherPossb(MultNumber, otherDigit, Digit) == 0)
{
return 0;//無解
}
printf("\nX-2:");
for(int i = 0; i < (int)strlen(otherDigit); i++)
{
printf("%c, ",otherDigit[i]);
}
if(parseSingleMedian(MultNumber, Digit, otherDigit) == 0)
{
return 0;//無解
}
printf("\nX-3:");
for(int i = 0; i< (int)strlen(Digit); i++)
{
printf("%c, ",Digit[i]);
}
printf("\nX-4:");
for(int i = 0; i< (int)strlen(otherDigit); i++)
{
printf("%c, ",otherDigit[i]);
}
tmp[0] = Digit[0];
Number *= atoi(tmp);
tmp[0] = otherDigit[0];
Number += atoi(tmp);
printf("\nX-Number:%d\n", Number);
//20008
int step = 4;
int intIndex;
for(int i = 1; i<Count/2; i++)
{
memset(otherDigit, 0, 11);
memset(Digit, 0, 11);
memset(endDigitList, 0, 11);
if(parseDigit(MultNumber, Digit) == 0)
{
isSuccess = false;
break;
}
printf("X-%d:",++step);
for(intIndex = 0; intIndex < (int)strlen(Digit); intIndex++)
{
printf("%c, ",Digit[intIndex]);
}
if(updataNoRepeatNumber(Number, Count, Digit, i) == 0)
{
isSuccess = false;
break;
}
printf("\nX-%d:", ++step);
for(intIndex = 0; intIndex < (int)strlen(Digit); intIndex++)
{
printf("%c, ",Digit[intIndex]);
}
if(parseOtherDigit(MultNumber, Number, Digit, otherDigit, i) == 0)
{
isSuccess = false;
break;
}
printf("\nX-%d:", ++step);
for(intIndex = 0; intIndex < (int)strlen(otherDigit); intIndex++)
{
printf("%c, ", otherDigit[intIndex]);
}
if(updataNoRepeatNumber(Number, Count, otherDigit, i) == 0)
{
isSuccess = false;
break;
}
printf("\nX-%d:", ++step);
for(intIndex = 0; intIndex < (int)strlen(otherDigit); intIndex++)
{
printf("%c, ", otherDigit[intIndex]);
}
tmp[0] = Digit[0];
Number += (atoi(tmp)* (int)pow(10, Count - i - 1));
tmp[0] = otherDigit[0];
Number += atoi(tmp) * (int)pow(10, i);
printf("\nX-Number:%d\n", Number);
}
//----------------part 1

redex 2007-09-11

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MARK.

xure1234 2007-09-11

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在VC中运行后得到结果：21978
//---------------------------------------------------------------------------
#include <iostream>
using namespace std;

int main(int argc, char* argv[])
{
int A = 2, B, C, D, E = 8;
long int temp1, temp2;

for(B = 0; B <= 9; B++)
{
if((B != A) && (B != E))
{
for(C = 0; C <= 9; C++)
{
if((C != A) && (C != B) && (C != E))
{
for(D = 0; D <= 9; D++)
{
if((D != A) && (D != B) && (D != C) && (D != E))
{
temp1 = A * 10000 + B * 1000 + C * 100 + D * 10 + E;
temp2 = E * 10000 + D * 1000 + C * 100 + B * 10 + A;
if((temp1 * 4) == temp2)
cout << temp1 << endl;
}
}
}
}
}

}
system("pause");

return 0;
}

gxdl5460 2007-09-11

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hustwei()
Too cool!

qianlima888 2007-09-11

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#include "stdafx.h"
#include <iostream>

using namespace std;

int main(void)
{
int a,b,c,d,e;
for(a=1;a<10;a++)
for(b=0;b<10;b++)
for(c=0;c<10;c++)
for(d=0;d<10;d++)
for(e=1;e<10;e++)
{
if (a==b || b==c || c==d || d==e)
{
continue;
}
if((a*10000+b*1000+c*100+d*10+e)*4==(e*10000+d*1000+c*100+b*10+a))
cout<<(a*10000+b*1000+c*100+d*10+e)<<"*4=="<<e*10000+d*1000+c*100+b*10+a<<endl;

}

system("pause");
return 0;
}
通过VS2005调试!

zhengwei1981 2007-09-10

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#include <iostream>
using namespace std;

int main()
{
for(int i=0;i<=9;i++)
{
for(int j=0;j<=9;j++)
{
for(int u=0;u<=9;u++)
{
for(int k=0;k<=9;k++)
{
for(int e=0;e<=9;e++)
{
if((i*10000+j*1000+u*100+k*10+e)*4==(e*10000+k*1000+u*100+j*10+i))
cout<<i<<endl<<j<<endl<<u<<endl<<k<<endl<<e<<endl;
}
}
}
}
}
cin.get();
}

ascent2006 2007-09-10

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LS的,为什么万位一定是2?

wlj040801 2007-09-10

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#include<iostream>
#include<cstdlib>
using namespace std;
int main()
{
int i,a,b,c,d,e;
for(i=20008;i<22500;i+=10)
{
a=2;/*万位一定是2*/
b=i/1000%10;
c=i/100%10%10;
d=i/10%10%10%10;
e=8;/*个位一定是8*/
if((b!=2&&c!=2&&d!=2&&b!=c&&b!=d&&b!=8&&c!=d&&c!=8&&d!=8&&i*4==8*10000+d*1000+c*100+b*10+2))

cout<<i<<endl;
}
system("pause");
return 0;
}

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求解：华为一道面试题

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