hpu1028 整数划分

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

#include<stdio.h>
#include<string.h>

int f[125][125];

int fun(int m,int n)
{
    if(f[m][n]) return f[m][n];
    if(m==1||n==1) return f[m][n]=1;
    if(n>m) return f[m][n]=fun(m,m);
    if(n==m) return f[m][n]=fun(m,n-1)+1;
    else return f[m][n]=fun(m,n-1)+fun(m-n,n);
}

int main()
{
    int n;
    while(scanf("%d",&n)!=EOF)
        printf("%d\n",fun(n,n));
    return 0;
}

//母函数代码

#include<stdio.h>
#include<string.h>

int a[125],b[125];

int main()
{
    int i,j,k,n;
    while(scanf("%d",&n)!=EOF)
    {
        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[j+k]+=a[j];
            for(j=0;j<=n;j++)
            {
                a[j]=b[j];
                b[j]=0;
            }
        }
        printf("%d\n",a[n]);
    }
    return 0;
}