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): 12606    Accepted Submission(s): 8903

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 代码
#include<stdio.h>#include<iostream>#define MAXN 310using namespace std;int main(){    int temp[MAXN],ans[MAXN];    int i,m=200,j,k,n;    for(i=0; i<=m; i++)    {        ans[i]=1;        temp[i]=0;    }    for(i=2; i<=m; i++)    {        for(j=0; j<=m; j++)            for(k=0; k+j<=m; k+=i)            {                temp[k+j] += ans[j] ;            }        for(j=0; j<=m; j++)        {            ans[j] = temp[j];            temp[j] = 0;        }    }    while(cin >> n&&n)    {        cout << ans[n]<< endl;;    }    return 0;}