(hdu1028)Ignatius and the Princess III

Problem Description

"Well, it seems the first problem is too easy. I will let you know how foolish you are later." feng5166 says.

"The second problem is, given an positive integer N, we define an equation like this:
  N=a[1]+a[2]+a[3]+...+a[m];
  a[i]>0,1<=m<=N;
My question is how many different equations you can find for a given N.
For example, assume N is 4, we can find:
  4 = 4;
  4 = 3 + 1;
  4 = 2 + 2;
  4 = 2 + 1 + 1;
  4 = 1 + 1 + 1 + 1;
so the result is 5 when N is 4. Note that "4 = 3 + 1" and "4 = 1 + 3" is the same in this problem. Now, you do it!"

 

 

Input

The input contains several test cases. Each test case contains a positive integer N(1<=N<=120) which is mentioned above. The input is terminated by the end of file.

 

 

Output

For each test case, you have to output a line contains an integer P which indicate the different equations you have found.

 

 

Sample Input

41020

 

 

Sample Output

542627

 题意是 给出一个数，问有多少种组合的和为这个数， 对于 5 = 1 + 4， 5 = 4 + 1属于同一种，显然的母函数。

#include <iostream>#include <cstring>#include <algorithm>using namespace std;int c1[121],c2[121];int main(){ int i,j,k; for(i = 0; i <= 120; i++) { c1[i] = 1; c2[i] = 0; } for(i = 2; i <= 120; i++) { for(j = 0; j <= 120; j++) { for(k = 0; k + j <= 120; k += i) { c2[j + k] += c1[j]; } } for(j = 0; j <= 120; j++) { c1[j] = c2[j]; c2[j] = 0; } } int n; while(cin >> n) { cout << c1[n] << endl; } return 0;}