HDOJ 1028 Ignatius and the Princess III (母函数)

本文介绍了一个编程问题“公主III”,该问题是计算给定正整数N的不同等式的数量。通过一个C++程序示例展示了如何使用动态规划解决这个问题,并提供了完整的代码实现。

 

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9532    Accepted Submission(s): 6722


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
 
 
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 
 5 using namespace std;
 6 
 7 const int maxn=200;
 8 
 9 int c1[maxn],c2[maxn];
10 
11 int main()
12 {
13     int n;
14 while(scanf("%d",&n)!=EOF)
15 {
16     for(int i=0;i<=n;i++)
17     {
18         c1[i]=1; c2[i]=0;
19     }
20 
21     for(int i=2;i<=n;i++)
22     {
23         for(int j=0;j<=n;j++)
24         {
25             for(int k=0;j+k<=n;k+=i)
26                 c2[k+j]+=c1[j];
27         }
28 
29         for(int j=0;j<=n;j++)
30         {
31             c1[j]=c2[j];  c2[j]=0;
32         }
33     }
34     printf("%d\n",c1[n]);
35 }
36 
37     return 0;
38 }

 

转载于:https://www.cnblogs.com/CKboss/p/3165204.html

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值