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杭电ACM 1028 Ignatius and the Princess III

neuxxm posted @ 2011年5月20日 09:57 in 杭电ACM 解题报告 with tags 母函数 , 2368 阅读

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4193 Accepted Submission(s): 2931

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

#include <stdio.h>
#include <string.h>
#define N 120
int c1[N+1], c2[N+1];
int main(){
    int i, j, k, n;
    for (i=0; i<=N; ++i){
        c1[i] = 1;
    }
    for (k=2; k<=N; ++k){
        memset(c2, 0, sizeof(c2));
        for (i=0; i<=N; ++i){
            for (j=i; j<=N; j+=k){
                c2[j] += c1[i];
            }
        }
        memcpy(c1, c2, sizeof(c2));
    }
    while (scanf("%d", &n) == 1){
        printf("%d\n", c1[n]);
    }
    return 0;
}

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