ACdream1431-Sum vs Product

Sum vs Product

Time Limit: 2000/1000MS (Java/Others) Memory Limit: 128000/64000KB (Java/Others)

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Problem

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

2
5

Sample Output

1
3

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

题意：让你求1~n中任意选n个数字，保证这n个数字加和与乘积相等，问有多少种方法。

解题思路：暴力搜索，加上剪枝。从2开始枚举，乘积总是比加和的上升速度快。那么如果前m个数的和加上剩下的n-m个1的值还是小于前m个数的乘积的话，那么就不可能有让最后的和跟积相等的情况了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <stack>

using namespace std;

int n,sum;

void dfs(int cur,int sum1,int mul,int k)
{
if(sum1+n-k<mul) return ;
else if(sum1+n-k==mul)
{
sum++;
return ;
}
for(int i=cur; i<=n; i++)
dfs(i,sum1+i,mul*i,k+1);
}

int main()
{
while(~scanf("%d",&n))
{
sum=0;
dfs(2,0,1,0);
printf("%d\n",sum);
}
return 0;
}