学步园

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4680    Accepted Submission(s): 3280

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

 

Author

Ignatius.L

 

 

 

母函数：

 

#include<stdio.h>
#include<iostream>
using namespace std;
const int MAXN=120;
int c1[MAXN+1],c2[MAXN+1];
void calc()
{
    int i,j,k;
    for(i=0;i<=MAXN;i++)
      {c1[i]=1;c2[i]=0;}
    for(k=2;k<=MAXN;k++)
    {
      for(i=0;i<=MAXN;i++)
        for(j=0;j+i<=MAXN;j+=k)
        {
            c2[i+j]+=c1[i];
        }  
      for(i=0;i<=MAXN;i++)
      {
          c1[i]=c2[i];
          c2[i]=0;
      }       
    }    
} 
int main()
{
    int n;
    calc();
    while(cin>>n)
    {
        cout<<c1[n]<<endl;
    }   
    return 0; 
}  

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
#define MAXN 121
int c1[MAXN],c2[MAXN];//c1存放目前所有函数的乘积，c2存放两个函数的临时乘积
int maxn;
void mufun(int n)
{
    int i,j,k;
    for(i=0;i<=n;i++)
    {
        c1[i]=1;
    }
    for(k=2;k<=n;k++)
    {
        for(i=0;i<=n;i++)
        {
            for(j=0;j+i<=n;j+=k)
            {
                c2[i+j]+=c1[i];
            }    
        }
        for(i=0;i<=n;i++)
        {
            c1[i]=c2[i];
            c2[i]=0;
        }        
    }         
}
int main()
{
    int n,i;
    while(scanf("%d",&n)!=EOF)
    //while(cin>>n[0]>>n[1]>>n[2])
    {
     
        memset(c1,0,sizeof(c1));
        memset(c2,0,sizeof(c2));
        mufun(n);
        printf("%d\n",c1[n]);   
    }
    return 0;    
}