# hdu 1028 Ignatius and the Princess III 【整数区划】

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hdu 1028 Ignatius and the Princess III 【整数划分】

# Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15730    Accepted Submission(s): 11092
链接： hdu 1028
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

#include <queue>
#include <vector>
#include <cstdio>
#include <string>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <algorithm>
using namespace std;
#define FIN             freopen("input.txt","r",stdin)
#define FOUT            freopen("output.txt","w",stdout)
#define CASE(T)         for(scanf("%d",&T);T--;)
const int maxn = 120 + 5;
int dp[maxn][maxn];
int dfs(int n, int m)
{
if(dp[n][m] != -1)      return dp[n][m];
if(n < 1 || m < 1)      return dp[n][m] = 0;
if(n == 1 || m == 1)    return dp[n][m] = 1;
if(n < m)               return dp[n][m] = dfs(n, n);
if(n == m)              return dp[n][m] = dfs(n, m - 1) + 1;
return dp[n][m] = dfs(n, m - 1) + dfs(n - m, m);
}
int main()
{
//    FIN;
int N;
while(~scanf("%d", &N))
{
printf("%d\n", dfs(N, N));
}
return 0;
}
```