poj 2718 切数问题 穷竭搜索

题意： 给一个已经排序号的数字，从中间切一刀，成两个数，要求这两个数的差最小

思路：暴力比较差最小值

1. stl中的next_permutation()函数进行排列  注意：这个函数必须从小到大才可以排序
2. 把所有排序写出，从中间切，前面一个数，后面一个数，找到最小

   if (a[mid])
   {
   int x = a[0]; int y = a[mid];
   for (int i = 1; i < mid; i++)
   x = x * 10 + a[i];
   for (int i = mid + 1; i < n; i++)
   y = y * 10 + a[i];
   if (ans > abs(x - y))
   ans = abs(x - y);
   }

解决问题的代码：

#include <iostream>
#include <math.h>
#include <algorithm>
#include <string.h>
#include <cstdio>
using namespace std;
int n, a[15];
void solve()
{
    while (a[0] == 0) next_permutation(a, a + n);
    int ans = 1000000;
    int mid = (n + 1) / 2;
    do {
        if (a[mid])
        {
            int x = a[0]; int y = a[mid];
            for (int i = 1; i < mid; i++)
                x = x * 10 + a[i];
            for (int i = mid + 1; i < n; i++)
                y = y * 10 + a[i];
            if (ans > abs(x - y))
                ans = abs(x - y);
        }
    } while (next_permutation(a, a + n));
    printf("%d
", ans);
    return;
}
int main()
{
    int t;
    char c;
    scanf("%d", &t);
    getchar();
    while (t--)
    {
        n = 0;
        memset(a, 0, sizeof(a));
        while ((c = getchar()) != '
')
        {
            if (c != ' ')
                a[n++] = c - '0';
        }
        if (n == 1) printf("%d
", a[0]);
        if (n == 2) printf("%d
", abs(a[1] - a[0]));
        else solve();
    }
    return 0;
}