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
Author
Ignatius.L
1 /* 2 母函数模板 3 http://blog.csdn.net/vsooda/article/details/7975485 详解 4 */ 5 #include6 #include 7 #define MAXN 121 8 9 using namespace std;10 11 int c1[MAXN];12 int c2[MAXN];13 14 int n;15 16 int main() {17 while(~scanf("%d",&n)) {18 for(int i=0;i<=n;i++) {19 c1[i]=1;c2[i]=0;20 }21 for(int i=2;i<=n;i++) {22 for(int j=0;j<=n;j++)23 for(int k=0;k+j<=n;k+=i)24 c2[j+k]+=c1[j];25 for(int j=0;j<=n;j++) {26 c1[j]=c2[j];27 c2[j]=0;28 }29 }30 printf("%d\n",c1[n]);31 }32 return 0;33 }