hdu1028Ignatius and the Princess III

"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!"

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.

For each test case, you have to output a line contains an integer P which indicate the different equations you have found.

4
10
20

5
42
627

Ignatius.L


dp[i][j]:用[1,j]里的数组成i的方法种数

dp[i][j]=dp[i][j-1]+dp[i-j][j]

由于dp[i][j-1]里组成i的方法中都是不包含j的

所以把j单独取出，保证所有方法至少含有一个j，

也就是dp[i-j][j]

#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#include<bitset>
#include<climits>
#include<list>
#include<iomanip>
#include<stack>
#include<set>
using namespace std;
int dp[200][200];
int dfs(int a,int b)
{
	if(dp[a][b]!=-1)
		return dp[a][b];
	if(b==1)
		return dp[a][b]=1;
	if(a<b)
		return dp[a][b]=dfs(a,a);
	if(a==b)
		return dp[a][b]=dfs(a,b-1)+1;
	return dp[a][b]=dfs(a,b-1)+dfs(a-b,b);
}
int main()
{
	int n;
	while(cin>>n)
	{
		memset(dp,-1,sizeof(dp));
		cout<<dfs(n,n)<<endl;
	}
}