HDU 1028 Ignatius and the Princess III（母函数模板）

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17839    Accepted Submission(s): 12508

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

 

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母函数初学者，还是看大神的总结吧http://blog.csdn.net/vsooda/article/details/7975485，下面是个模板题。。

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int main(){int n,i,j,k;int c1[140];//保存每个数字可以组合的数目 int c2[140];//中间量，保存每一次的情况 while(scanf("%d",&n)!=EOF){for(i=0;i<=n;i++)//初始化 {c1[i]=1;c2[i]=0;}for(i=2;i<=n;i++) {for(j=0;j<=n;j++){for(k=0;k+j<=n;k+=i)c2[k+j]+=c1[j];}for(j=0;j<=n;j++){c1[j]=c2[j];c2[j]=0; } }printf("%d\n",c1[n]);}return 0;}