整数划分(hdu 1028)

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17582    Accepted Submission(s): 12324

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

#include <stdio.h>#include <string.h>#include <stdlib.h>int main(){    int dp[150][150];    int n;    while(~scanf("%d", &n)){        int i, j;        for(i = 1;i <= n;i++){            dp[i][1] = dp[1][i] = dp[0][i] = 1;        }        for(i = 2;i <= n;i++){            for(j = 2;j <= n;j++){                if(j <= i){                    dp[i][j] = dp[i][j-1] + dp[i-j][j];                }else {                    dp[i][j] = dp[i][i];                }            }        }        printf("%d\n", dp[n][n]);    }    return 0;}