acdream 1431 Sum vs Product

Sum vs Product

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

SubmitStatisticNext
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

题解及代码：

通过打表前几项我们会发现构成n。比方n=5时。其形式之中的一个是1 1 2 2 2，都是这样的非常多1，然后其它数字组合的形式。那么我们就能够枚举除了1以外的数字的组合，来计算sum[n]。比方数字组合为2 3 4，那么依据公式我们知道2*3*4=24，2+3+4=9，那么我们还须要补上15个1，加上2 3 4 这三个数字，总共是18个数字，那么2 3 4必定属于sum[18]里面的一中情况。得到验证。这样我们就能用dfs来求出全部的情况数了。

以下的代码是dfs的代码，由于怕超时的缘故，题目AC的代码是打表之后交的。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
int sum[510];
void init()
{
  memset(sum,0,sizeof(sum));
}
void dfs(int nt,int nu,int su,int k)
{
    for(int i=k;i<=500;i++)
    {
        if(nu*i>1000) break;
        sum[nu*i-su-i+nt+1]++;
        //printf("%d %d %d %d %d\n",nu,su,i,nt+1,nu*i-su-i+nt+1);
        dfs(nt+1,nu*i,su+i,i);
    }
}

int main()
{
    init();
    for(int i=2;i<=500;i++)
    dfs(1,i,i,i);
    for(int i=2;i<=500;i++)
        printf("%d,",sum[i]);
    return 0;
}

版权声明：本文博客原创文章。博客，未经同意，不得转载。