杭电1028 Ignatius and the Princess III（整数拆分）

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

 

 

Author

Ignatius.L

 

    题目描述：把一个整数拆分成一个或多个的整数之和，问总共有多少种拆分方法？

    问题的解决：多个整数可以是0,1,2,3、、、n，设这些整数为i，则所需要i的最多的个数为 n/i；母函数写法

 

#include <stdio.h>
#include <string.h>

#define MAX_INDEX 120
int main()
{
    int n, i, j, k;
    //母函数所需的两个一维数组
    int c1[MAX_INDEX+1], c2[MAX_INDEX+1];
    while( scanf( "%d",&n ) != EOF )
    {
        memset(c1,0,sizeof(c1));
        memset(c2,0,sizeof(c2));
        for( i = 0; i <= n; i++  )
            c1[i] = 1;
        for( i = 2; i <= n; i++ )
        {
            for( j = 0; j <= n; j++ )
                for( k = 0; k + j <= n; k += i )
                    c2[k+j] += c1[j];
            for( j = 0; j <= n; j++ )
            {
                c1[j] = c2[j];
                c2[j] = 0;
            }
        }
        printf( "%d\n", c1[n] );
    }
    return 0;
}