Ignatius and the Princess III

Ignatius and the Princess III

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10312    Accepted Submission(s): 7318

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

思路：一开始拿到这个题目以为是找规律，有递推关系什么的，最后找了好长时间没找到规律，上网查了一下才发现是用母函数做，就是把数的加法和指数乘法的幂的加法联系起来，母函数：G(x)=(1+x+x^2+x^3+x^4+.....)*(1+x^2+x^4+x^6+....)*(1+x^3+x^6+x^9+....)*... ,x^n的系数就是n的拆分方案数！其实这个不难理解，因为x^n的系数是多少就表明有多少个x^n相加得来，换句话说就是有多少种x的幂之和的拼凑方案，即本题所求。

#include<stdio.h>
int a[],b[];   //　a[i]表示x^i的系数，为临时值，b[i]表示x^i的系数，为最终值；
int main()
{
    int i,j,k,n;
    for(i =;i <=;i ++)
    {
        a[i] =;
        b[i] =;
    }
    for(i =;i <=;i ++)
    {
        for(j =;j <=;j ++)
        {
            for(k =;k+j <=; k += i)
                a[k+j] += b[j];      //因为x^(k+j)是从x^j得来的，故它的系数应该在原有系数的数值的基础上加上x^j　　　　　　　　　　　　　　　　　
　　　　　　　　　　　　　　　　　　　　　　的系数（这是关键的重点！！！这就是为什么我们要用两个数组的目的）
        }
        for(j =;j <=;j ++)
        {
            b[j] = a[j];
            a[j] =;
        }
    }
    while(~scanf("%d",&n))
        printf("%d\n",b[n]);
　　return 0;
}