HDU 1028-Ignatius and the Princess III(经典母函数模型)

题目传送门

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21747    Accepted Submission(s): 15182

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
//经典母函数#include <stdio.h>#define maxn 125int c1[maxn];int c2[maxn];int main(){for(int i=0; i<maxn; i++){c1[i] = 1;c2[i] = 0;}for(int i=2; i<maxn; i++){for(int j=0; j<maxn; j++)for(int k=0; k+j<maxn; k+=i){c2[j+k] += c1[j];}for(int j=0; j<maxn; j++){c1[j] = c2[j];c2[j] = 0;}}int n;while(~scanf("%d", &n)){printf("%d\n", c1[n]);}return 0;}

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