母函數--Ignatius and the Princess III

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25404    Accepted Submission(s): 17555

 

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

 

 

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 思路：直接上母函數即可

#include<stdio.h>
#define MAX 150 

int c1[MAX],c2[MAX],n;

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