hdu acm 1028 数字拆分Ignatius and the Princess III

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 ;
 }