hdu 1028 Sample Ignatius and the Princess III (母函数)

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25929    Accepted Submission(s): 17918

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
4
10
20
 

Sample Output
5
42
627

C/C++:

#include <map>
#include <queue>
#include <cmath>
#include <vector>
#include <string>
#include <cstdio>
#include <cstring>
#include <climits>
#include <iostream>
#include <algorithm>
#define INF 0x3f3f3f3f
#define LL long long
using namespace std;
const int MAX = 2e2 + 10;

int n, ans[MAX], temp[MAX];

void calc()
{
    for (int i = 0; i <= 130; ++ i)
        ans[i] = 1, temp[i] = 0;
    for (int i = 2; i <= 130; ++ i)
    {
        for (int j = 0; j <= 130; ++ j)
            for (int k = 0; j + k <= 130; k += i)
                temp[j + k] += ans[j];
        for (int j = 0; j <= 130; ++ j)
            ans[j] = temp[j], temp[j] = 0;
    }
}

int main()
{
    calc();
    while (~scanf("%d", &n))
        printf("%d
", ans[n]);
    return 0;
}