整数划分(hdu 1028)

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本文介绍了一个编程问题“公主III”，该问题是关于给定正整数N的不同等式的数量计算。通过动态规划的方法解决了这个问题，并提供了完整的C语言实现代码。

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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

4
10
20

Sample Output

5
42
627

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;
}