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): 18839 Accepted Submission(s): 13226

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

母函数，求组合排列数

#include <iostream>#include <cstdio>#include <cstring>#define inf 0x6f6f6f6fusing namespace std;int main(){    int c1[125];    int c2[125];    int n,i,j,k;    while(cin>>n)    {        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[j+k]+=c1[j];                }            }            for(j=0; j<=n; j++)            {                c1[j]=c2[j];                c2[j]=0;            }        }        cout<<c1[n]<<endl;    }    return 0;}