POJ

Smallest Difference

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11665 Accepted: 3155

Description

Given a number of distinct decimal digits, you can form one integer by choosing a non-empty subset of these digits and writing them in some order. The remaining digits can be written down in some order to form a second integer. Unless the resulting integer is 0, the integer may not start with the digit 0. 

For example, if you are given the digits 0, 1, 2, 4, 6 and 7, you can write the pair of integers 10 and 2467. Of course, there are many ways to form such pairs of integers: 210 and 764, 204 and 176, etc. The absolute value of the difference between the integers in the last pair is 28, and it turns out that no other pair formed by the rules above can achieve a smaller difference.

Input

The first line of input contains the number of cases to follow. For each case, there is one line of input containing at least two but no more than 10 decimal digits. (The decimal digits are 0, 1, ..., 9.) No digit appears more than once in one line of the input. The digits will appear in increasing order, separated by exactly one blank space.

Output

For each test case, write on a single line the smallest absolute difference of two integers that can be written from the given digits as described by the rules above.

Sample Input

10 1 2 4 6 7

Sample Output

28

题意：

        输入几个数字（0~9），随机组合成两个数字，使其差最小。

题解：

        全排列。分为两种情况。第一种是总共有2个数字，那么0可以作为单独的一个数字。第二种是有多于2个数字，那么0就不能作为单独的数字出现，要判断0是否为数字的首位的情况。要使差最小，两个数字的位数一定是相同或只差一位。那么就可以直接在全排列函数内将a数组从中间划开，分为两个数字。

#include<stdio.h>#include<string.h>#include<algorithm>#include<ctype.h>#include<math.h>#include<stdlib.h>using namespace std;int a[20],cnt;int main(){    int t;    scanf("%d",&t);    getchar();    while(t--)    {        char str[50];        cnt=0;        gets(str);        for(int i=0;str[i];i++)            if(isdigit(str[i]))                a[cnt++]=str[i]-'0';        sort(a,a+cnt);        int minn=99999999;        do        {            int first=0,second=0,flag=0;            if(a[0]==0 || a[cnt/2]==0 && cnt>2)                continue;            for(int i=0;i<cnt/2;i++)                first=first*10+a[i];            for(int j=cnt/2;j<cnt;j++)                second=second*10+a[j];            minn=min(minn,abs(first-second));        }while(next_permutation(a,a+cnt));        printf("%d\n",minn);    }    return 0;}