HDU 1028 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): 9408    Accepted Submission(s): 6642

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

 

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1028

 

题目分析：题目要求的是整数的拆分方案个数，用母函数处理。

 

母函数详解链接：http://www.wutianqi.com/?p=596

 

#include <stdio.h>

int N;
const int MAX_NUM = 125;
int   solNum[MAX_NUM];
int   temp[MAX_NUM];

void InitGenerFun()
{
    for(int i = 0; i < MAX_NUM; i++)
    {
        solNum[i] = 1;
        temp[i]   = 0;
    }
    for(int i = 2; i < MAX_NUM; i++)
    {
        for(int j = 0; j < MAX_NUM; j++)
        {
            for(int k = 0; k+j < MAX_NUM; k+=i)
            {
                temp[k+j] += solNum[j];
            }
        }
        for(int k = 0; k < MAX_NUM; k++)
        {
            solNum[k] = temp[k];
            temp[k]   = 0;
        }
    }
}

int main()
{
    InitGenerFun();

    while(scanf("%d", &N) != EOF)
    {
        printf("%d\n", solNum[N]);
    }

    return 0;
}