hdu acm 1028 数字拆分Ignatius and the Princess III

时间：2014-05-01 11:48:06 收藏：0 阅读：310

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

 1 /*
 2 hdu acm 1028  数字拆分，母函数模板题
 3 by zhh
 4 */
 5 #include <iostream>
 6 #include <stdio.h>
 7 #include <stdlib.h>
 8 #include <algorithm>
 9 #include <cstring>
10 
11 using namespace std;
12 #define maxx 120
13 int ans[maxx+2],temp[maxx+2];
14 void init()//母函数打表
15 {
16     for(int i=0;i<=maxx;i++)//初始化第一个式子系数
17     {
18         ans[i]=1;
19         temp[i]=0;//用于临时保存每次相乘的结果
20     }
21     for(int i=2;i<=maxx;i++)//循环每一个式子
22     {
23         for(int j=0;j<=maxx;j++)//循环第一个式子各项
24             for(int k=0;k+j<=maxx;k+=i)//下个式子的各项
25                 temp[k+j]+=ans[j];//结果保存到temp数组中
26         for(int j=0;j<=maxx;j++)//临时保存的值存入ans数组
27         {
28             ans[j]=temp[j];
29             temp[j]=0;
30         }
31     }
32 }
33 int main()
34 {

35     init();
36     int n;
37     while(scanf("%d",&n)!=EOF)
38     {
39         cout<<ans[n]<<endl;
40     }
41     return 0;
42 }

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