ACdream 1431 Sum vs Product

题目链接：http://115.28.76.232/problem?

pid=1431

Sum vs Product

Time Limit: 4000/2000MS (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

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这个题目事实上仅仅要枚举下因子。暴力dfs一下就能过。。

。

//#pragma comment(linker, "/STACK:36777216")
#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <climits>
#include <cassert>
#include <complex>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
#define CLR(A,B) memset(A,B,sizeof(A))
#define CPY(A,B) memcpy(A,B,sizeof(B))
#define long long ll
#define MP(A,B) make_pair(A,B)
int N;
long long cnt;
int dfs(int now,long long sum,long long mul,int pre){
if(now==N){
cnt++;
return 1;
}
for(int i=pre;i<2000;i++){
int ts=sum+i,tm=mul*i;
int ms=ts+N-now-1,mm=tm;
if(ms==mm){cnt++;continue;}
if(ms<mm) return 0;
dfs(now+1,ts,tm,i);
}
}
int main(){
int F[550];
for(int i=2;i<=500;i++){
N=i;cnt=0;
dfs(0,0,1,2);
F[i]=cnt;
}
while(~scanf("%d",&N)) printf("%d\n",F
);
return 0;
}