HDU 1028 (dp)

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 16780 Accepted Submission(s): 11808



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

首先给出一个递归函数fun(int n,int m);
n为需要划分的数，m为最大的加数，比如 4 = 1 + 1 + 1＋１　　　ｎ＝４，ｍ＝４

１、当ｎ　＞　ｍ的时候　就是上面所说的情况　ｆｕｎ（ｎ，ｍ）　＝　ｆｕｎ（ｎ，ｍ－１）＋ｆｕｎ（ｎ－ｍ，ｍ）
　　　　　　　　　　　　　　　　　　　　　　　ｆｕｎ（４，４）　＝　ｆｕｎ（４，３）＋ｆｕｎ（１，３）就是再递归去求ｎ＝４，ｍ＝３，
２、当　ｎ　＝＝　ｍ　时．等于１＋ｆｕｎ（ｎ，ｍ－１）
３、当　ｎ　＜　ｍ　时，等于　ｆｕｎ（ｎ，ｎ）

说的不清，看下面的图

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;

int main()
{
int a[121][121],i,j,k,m,n,t;

for(i=1;i<=120;i++)
{
a[i][1] = 1;
a[1][i] = 1;
}

for(i=2;i<=120;i++)
{
for(j=2;j<=120;j++)
{
if(i > j)
a[i][j] = a[i][j-1] + a[i-j][j];
else
if(i == j)
a[i][j] = a[i][j-1] + 1;
else
a[i][j] = a[i][i];
}
}

while(cin>>n)
{
cout<<a

<<endl;
}
}