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): 9848    Accepted Submission(s): 6962

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

 

Author

Ignatius.L

题意：拆分整数

题解：构造母函数G（x）=（1+x+……）（1+x^2+……）……即可

#include<stdio.h>int dp[128],temp[128];int main(){    int i,j,k,n;    while(scanf("%d",&n)>0)    {        for(i=0;i<=n;i++) dp[i]=1,temp[i]=0;        for(i=2;i<=n;i++)        {            for(j=0;j<=n;j++)            {                for(k=0;k+j<=n;k+=i)                    temp[j+k]+=dp[j];            }            for(j=0;j<=n;j++) dp[j]=temp[j],temp[j]=0;        }        printf("%d\n",dp[n]);    }    return 0;}