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): 9017    Accepted Submission(s): 6341

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

 

 

 

#include<iostream>

#include<cstdio>

#include<cstring>



using namespace std;



int c1[130],c2[130];



int main(){



    //freopen("input.txt","r",stdin);



    int n;

    while(~scanf("%d",&n)){

        for(int i=0;i<n;i++){

            c1[i]=0;

            c2[i]=0;

        }

        c1[0]=1;

        for(int i=1;i<=n;i++){      //要乘以n个多项式

            for(int j=0;j<=n;j++)   //c1的各项的指数

                for(int k=0;j+k*i<=n;k++)   //k*i表示被乘多项式各项的指数,(X^0*i + X^1*i + X^2*i + ……)

                    c2[j+k*i]+=c1[j];       //指数相加得j+k*i,加多少只取决于c1[j]的系数，因为被乘多项式的各项系数均为1

            for(int j=0;j<=n;j++){      

                c1[j]=c2[j];

                c2[j]=0;

            }

        }

        printf("%d\n",c1[n]);

    }

    return 0;

}