hdu 1028 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): 9937 Accepted Submission(s): 7032

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

41020

Sample Output

542627

Author

Ignatius.L

今天打了打篮球，好累~~~~~~~~~~~

code:

#include<iostream>using namespace std;int dp[121][121];int main(){    int i,j;     int n;    for (i=1;i<=121;i++)dp[1][i]=dp[i][1]=1;    for (i=2;i<121;i++)    {        for (j=2;j<=121;j++)        {            if (i<j) dp[i][j]=dp[i][i];            else if (i==j)dp[i][j]=1+dp[i][j-1];            else if (i>j) dp[i][j]=dp[i-j][j]+dp[i][j-1];        }    }       while (cin>>n)cout<<dp[n][n]<<endl;    return 0;}