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HDU 1028 简单动态规划

2024-08-11 05:43:32   211人阅读

"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

41020

Sample Output

542627
分治或者动态规划吧
给定一个正整数n，求它有多少种正整数和a1、a2……的组合形式。
m为这些正整数中最大的，有如下方程
f(n,m) = 1---------------n=1/m=1
         f(n-m,m)+f(n,m-1)--m<=n
         f(n,n)-------------m>n
         1------------------n=0
技术分享
 1 #include <cstdio> 2 #include <cstring> 3 #include <iostream> 4 #include <algorithm> 5 #include <vector> 6 #include <queue> 7 #include <set> 8 #include <map> 9 #include <string>10 #include <cmath>11 #include <stdlib.h>12 #define MAXSIZE 12513 using namespace std;14 15 int n;16 int f[MAXSIZE][MAXSIZE];17 int main()18 {19     //freopen("caicai.txt","r",stdin);20     int i,j;21     for(i = 1;i<MAXSIZE;i++)22     {23         f[i][1] = 1;24         f[1][i] = 1;25         f[0][i] = 1;//下文可能出现i-j=026     }27     for(i = 2;i<MAXSIZE;i++)28         for(j = 2;j<MAXSIZE;j++)29         {30             if(i>=j)31                 f[i][j] = f[i-j][j]+f[i][j-1];32             else33                 f[i][j] = f[i][i];34         }35     while(scanf("%d",&n)!=EOF)36     {37         cout<<f[n][n]<<endl;38     }39     return 0;40 }
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HDU 1028 简单动态规划


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