婚恋网站模板浙江义乌外发加工网

 思路：1.深度枚举所有排列情况

            2.设置为每个排列设置两个断点，分为三部分：a,b,c

            3.转换为乘法判断条件，满足加一

代码如下：（可用next_permutation全排列函数代替dfs） 

#include<iostream>
#include<stdio.h>
using namespace std;
int ans = 0;int visit[11] = { 0 };int arr[11] = { 0 }, k = 0;int tonum(int l, int r) {//将num数组的第l至r个数转为一个数int re = 0;for (int i = l; i <r; i++) {re = re * 10 + arr[i];}return re;
}void dec();void dfs();int l = 0;int main()
{scanf("%d", &l);dfs();cout << ans << endl;return 0;
}
void dfs()
{for (int i = 1; i <= 9; i++){if (!visit[i]){visit[i] = 1;arr[k++] = i;dfs();if (k == 9)		dec();k--;visit[i] = 0;}}
}void dec()
{for (int i = 1; i <= 7; i++){	int a = tonum(0, i);for (int j = i; j <= 8; j++){int b = tonum(i, j);int c = tonum(j, 9);if ((l - a) * c == b) {ans++;}}}
}