[1] strong %d fuck any build other main \n

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25867 Accepted Submission(s): 17879

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 G(x) = (1 + x^1 + x^2 + x^3 ......) * (1 + x^2 + x^4 + x^8.....) * (1 + x^3 + x^6 + x^9 ......)........... x^k的系數即為數字k的拆分方案數

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 int c1[125],c2[125];
 4 int n;
 5 
 6 int build(){
 7     for
 
 (int i = 0;i <= 120;++i){
 8         c1[i] = 1,c2[i] = 0;
 9     }
10     for (int i = 2;i <= 120;++i){
11         if (c1[i] == 0) return 0 * printf("%d fuck",i);
12         for (int j = 0;j <= 120;++j){
13             for (int k = 0;k+j <= 120;k+=i){
14                     c2[k+j] += c1[j];
15             }
16         }
17         for (int k = 0;k <= 120;++k){
18                 c1[k] = c2[k];
19                 c2[k] = 0;
20         }
21     }
22 }
23 
24 
25 int main(){
26     build();
27     while(~scanf("%d",&n)){
28         printf("%d\n",c1[n]);
29     }
30 }

HDU1028 Ignatius and the Princess III 母函數