acdream 1431 Sum vs Product

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

題解及代碼：

通過打表前幾項我們會發現構成n，比如n=5時，其形式之一是1 1 2 2 2，都是這種很多1，然後其他數字組合的形式。那麼我們就可以枚舉除了1以外的數字的組合，來計算sum[n]。比如數字組合為2 3 4，那麼根據公式我們知道2*3*4=24，2+3+4=9，那麼我們還需要補上15個1，加上2 3 4 這三個數字，總共是18個數字，那麼2 3 4必然屬於sum[18]裡面的一中情況。得到驗證，這樣我們就能用dfs來求出所有的情況數了。

下面的代碼是dfs的代碼，因為怕超時的緣故，題目AC的代碼是打表之後交的。

#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;
}