ACdream 1431(爆搜)

题目链接：http://acdream.info/problem?pid=1431

Sum vs Product

Time Limit: 4000/2000MS (Java/Others)Memory Limit: 128000/64000KB (Java/Others)

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Problem Description

      Peter has just learned mathematics. He learned how to add, and how to multiply. The fact that 2 + 2 = 2 × 2 has amazed him greatly. Now he wants find more such examples. Peters calls a collection of numbers beautiful if the product of the numbers in it is equal to their sum. 

      For example, the collections {2, 2}, {5}, {1, 2, 3} are beautiful, but {2, 3} is not. 

      Given n, Peter wants to find the number of beautiful collections with n numbers. Help him!

Input

      The first line of the input file contains n (2 ≤ n ≤ 500)

Output

      Output one number — the number of the beautiful collections with n numbers.

Sample Input

25

Sample Output

13

Hint

The collections in the last example are: {1, 1, 1, 2, 5}, {1, 1, 1, 3, 3} and {1, 1, 2, 2, 2}.

Source

Andrew Stankevich Contest 23

Manager

mathlover

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思路：比赛时想了好久，还打表找出10以内的所以解，但是最后还是没有做出来；

看了题解有所悟，暴力搜索；

直接上代码：

#include <iostream>#include <stdio.h>#include <string>#include <string.h>#include <cstdio>#include <cmath>using namespace std;int n,cnt;void dfs(int now,int N,int sum,int mul){ if(now==n+1) {   if(sum==mul)cnt++;   return; } for(int i=1;i<=N;i++) {   if(sum+i+n-now<mul*i)return;   dfs(now+1,i,sum+i,mul*i); }}void solve(){ cnt=0; dfs(1,n,0,1); //从第一层开始搜 printf("%d\n",cnt);}int main(){  while(scanf("%d",&n)!=EOF)  {    solve();  }  return 0;}