Sum vs Product

Time Limit: 4000/2000MS (Java/Others)

Memory Limit: 128000/64000KB (Java/Others)

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

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 
     
      #include 
      
       #include 
       
        #include 
        
         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;}
        
       
      
     

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