poj 2718 Smallest Difference（暴力）

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

1
0 1 2 4 6 7

Sample Output

28

Source

Rocky Mountain 2005

题意：给出不超过十个1位数，问这些数排列组合出的两个前导不为零数差的绝对值最小是多少？

思路：暴力枚举这些1位数的所有排列，用next_permutation函数是非常方便的。因为分成了两部分，所以从排列的中间进行组合再做差得到最后的结果。因为前导不是零，所以要判断当两个数前导为零时continue。

AC代码：

#include <iostream>
//#include <bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

int main()
{
    char s[3000];
    int a[20],n;
    while(~scanf("%d",&n))
    {
        getchar();
        while(n--)
        {
            gets(s);
            int len=strlen(s);
            int k=0;
            for(int i=0; i<len; i=i+2)
            {
                if(s[i]>='0'&&s[i]<='9')
                {
                    a[k++]=s[i]-'0';
                }
            }
            int    _min=999999;
            int ans1=0,ans2=0;
            sort(a,a+k);
            do
            {
                ans1=0,ans2=0;
                if(!a[0]||!a[k/2]&&k>2)
                continue;
                    for(int i=0; i<k/2; i++)
                        ans1=ans1*10+a[i];
                    for(int i=k/2; i<k; i++)
                        ans2=ans2*10+a[i];
                    if(abs(ans1-ans2)<_min)
                        _min=abs(ans1-ans2);
                }while(next_permutation(a,a+k));
            printf("%d\n",_min);
        }
    }
    return 0;
}