HDU1028

Ignatius and the Princess III

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
//HDU1028基础题//Ignatius and the Princess III//2017.05.26 by wyj#includeusing namespace std;int P[121][121];int main(){for (int i = 1;i <= 120;i++){for (int j = 1;j <= 120;j++){if (i == 1 || j == 1)P[i][j] = 1;else if (i < j)P[i][j] = P[i][i];else if (i == j)P[i][j] = P[i][j - 1] + 1;elseP[i][j] = P[i][j - 1] + P[i - j][j];}}int n;while (cin >> n){cout << P[n][n] << endl;}return 0;}