【hdoj】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


[align=left]Problem Description[/align] “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!”

[align=left]Input[/align] 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.

[align=left]Output[/align] For each test case, you have to output a line contains an integer P which indicate the different equations you have found.

[align=left]Sample Input[/align]

4

10

20

[align=left]Sample Output[/align]

5

42

627

#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
<< 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