hdu-1028-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): 12791 Accepted Submission(s): 9047

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

41020

Sample Output

542627
母函数的运用
#include<iostream>using namespace std;int a[125],b[125];int main(){    int n,i,j,k;    while(cin>>n)    {        for(i=0;i<=n;i++)        {            a[i]=1;            b[i]=0;        }        for(i=2;i<=n;i++)        {            for(j=0;j<=n;j++)            for(k=0;k+j<=n;k+=i)            b[k+j]+=a[j];            for(j=0;j<=n;j++)            {                a[j]=b[j];                b[j]=0;            }        }       cout<<a[n]<<endl;    }    return 0;}