Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 12521    Accepted Submission(s): 8838

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

整数拆分无限取，跟着包子做的题，就当做模板来用吧。

#include <stdio.h>
#define maxn 122

int c1[maxn], c2[maxn];

int main()
{
    int n, i, j, k;
    while(scanf("%d", &n) != EOF){
        for(i = 0; i <= n; ++i){
            c1[i] = 1; c2[i] = 0;
        }
        for(i = 2; i <= n; ++i){
            for(j = 0; j <= n; ++j)
                for(k = j; k <= n; k += i)
                    c2[k] += c1[j];
            for(k = 0; k <= n; ++k){
                c1[k] = c2[k]; c2[k] = 0;
            }
        }
        printf("%d\n", c1[n]);
    }
    return 0;
}

HDU1028 Ignatius and the Princess III 【母函数模板题】