hdu 1028 Ignatius and the Princess III(分解正整数的方案数)

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19270    Accepted Submission(s): 13527

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

 

题意：分解n有多少种方案

思路：模板题

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>using namespace std;int main(){    int a[350],b[350],i,j,k,n;    while(cin>>n&&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;}