Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11810    Accepted Submission(s): 8362

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
 

最简单的母函数模板题,用于学习和回顾母函数非常方便,代码也可直接做模板使用

 /*
hdu acm 1028 数字拆分,母函数模板题
by zhh
*/
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
#include <cstring> using namespace std;
#define maxx 120
int ans[maxx+],temp[maxx+];
void init()//母函数打表
{
for(int i=;i<=maxx;i++)//初始化第一个式子系数
{
ans[i]=;
temp[i]=;//用于临时保存每次相乘的结果
}
for(int i=;i<=maxx;i++)//循环每一个式子
{
for(int j=;j<=maxx;j++)//循环第一个式子各项
for(int k=;k+j<=maxx;k+=i)//下个式子的各项
temp[k+j]+=ans[j];//结果保存到temp数组中
for(int j=;j<=maxx;j++)//临时保存的值存入ans数组
{
ans[j]=temp[j];
temp[j]=;
}
}
}
int main()
{
init();
int n;
while(scanf("%d",&n)!=EOF)
{
cout<<ans[n]<<endl;
}
return ;
}

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