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

Code Display

/* *HDOJ_1028 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 */ #include<iostream> using namespace std; const int lmax = 120; int c1[lmax + 1], c2[lmax + 1]; int main() { int n, i, j, k; while (cin >> n) { for (i = 0; i <= n; i++) { c1[i] = 1; c2[i] = 0; } for (i = 2; i <= n; i++) { for(j=0;j<=n;j++) for (k = 0; k + j <= n; k += i) { c2[j + k] += c1[j]; } for (j = 0; j <= n; j++) { c1[j] = c2[j]; c2[j] = 0; } } cout << c1[n] << endl; } return 0; }

结果