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，比如n=5时，其形式之一是1 1 2 2 2，都是这种很多1，然后其他数字组合的形式。那么我们就可以枚举除了1以外的数字的组合，来计算sum[n]。比如数字组合为2 3 4，那么根据公式我们知道2*3*4=24，2+3+4=9，那么我们还需要补上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必然属于sum[18]里面的一中情况。得到验证，这样我们就能用dfs来求出所有的情况数了。
下面的代码是dfs的代码，因为怕超时的缘故，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; }