HDU 1028 简单动态规划

"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
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

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <stdlib.h>
#define MAXSIZE 125
using namespace std;

int n;
int f[MAXSIZE][MAXSIZE];
int main()
{
//freopen("caicai.txt","r",stdin);
int i,j;
for(i = 1;i<MAXSIZE;i++)
{
f[i][1] = 1;
f[1][i] = 1;
f[0][i] = 1;//下文可能出现i-j=0
}
for(i = 2;i<MAXSIZE;i++)
for(j = 2;j<MAXSIZE;j++)
{
if(i>=j)
f[i][j] = f[i-j][j]+f[i][j-1];
else
f[i][j] = f[i][i];
}
while(scanf("%d",&n)!=EOF)
{
cout<<f

<<endl;
}
return 0;
}

View Code