Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16028    Accepted Submission(s): 11302

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

题解：相当于取砝码，每个砝码可以取多次；

4=1+1+1+1=1+1+2=1+3=2+2；

则母函数为：

相当于1的砝码取法从0....n,2的砝码0....n/2。。。。。。k的砝码从0....n/k；

代码：

 1 #include<stdio.h>
 2 const int MAXN=10010;
 3 int main(){
 4     int a[MAXN],b[MAXN],N;
 5     while(~scanf("%d",&N)){
 6         int i,j,k;
 7         for(i=0;i<=N;i++){
 8             a[i]=1;b[i]=0;
 9         }
10         for(i=2;i<=N;i++){
11             for(j=0;j<=N;j++)
12                 for(k=0;k+j<=N;k+=i)
13                     b[j+k]+=a[j];
14                 for(j=0;j<=N;j++)
15                     a[j]=b[j],b[j]=0;
16             }
17         printf("%d\n",a[N]);
18     }
19     return 0;
20 }