hdu acm 1028 Ignatius and the Princess III

source : http://acm.hdu.edu.cn/showproblem.php?pid=1028

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19203    Accepted Submission(s): 13486

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

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#include <iostream>#include <string.h>#include <cstdio>using namespace std;int main(){   int dp[200];   int n = 0, i = 0, j = 0;   memset(dp, 0, sizeof(dp));   dp[0] = 1;// 1 = 1    // list   for(i = 1;i < 121;i++){        for(j = i;j < 121;j++){            dp[j] += dp[j - i];        }   }   while(scanf("%d", &n) != EOF){        cout << dp[n] << endl;   }   return 0;}//问题分析： DP动态规划//      寻找最小划分每一个问题的子问题， 因此得出题的解

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