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+2],temp[maxx+2];
void init()//母函数打表
{
    for(int i=0;i<=maxx;i++)//初始化第一个式子系数
    {
        ans[i]=1;
        temp[i]=0;//用于临时保存每次相乘的结果
    }
    for(int i=2;i<=maxx;i++)//循环每一个式子
    {
        for(int j=0;j<=maxx;j++)//循环第一个式子各项
            for(int k=0;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=0;j<=maxx;j++)//临时保存的值存入ans数组
        {
            ans[j]=temp[j];
            temp[j]=0;
        }
    }
}
int main()
{
    init();
    int n;
    while(scanf("%d",&n)!=EOF)
    {
        cout<<ans[n]<<endl;
    }
    return 0;
}