Ignatius and the Princess III（母函数/生成函数）

题面

"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!"

是个组合拆分问题，去查了一下，有个母函数/生成函数，将加变成了乘，即移到指数上，两项相乘，指数相加

 一定要落实到代码上，模拟乘法

#include 
using namespace std;
const int N = 1e5 + 10;
int a[N], b[N];//a存对应指数i下的系数， b用来过度当中间值
int n;
int main()
{
    while(~scanf("%d", &n))
    {
        for (int i = 0; i <= n; i ++ )//用第一个括号式初始一下每个项的值
        {
            a[i] = 1;
        }
        for (int i = 2; i <= n; i ++ )//枚举第几个括号式（2~n）
        {
            for (int j = 0; j <= n; j ++ )//j代表此时指数的从小到大的项
            {
                for (int k = 0; k + j <= n; k += i)//k代表指数（k每次+i是第i个括号的递增）
                {
                    b[k + j] += a[j];//对应乘起来变成指数k+j
                }
            }
            for (int j = 0; j <= n; j ++ )
            {
                a[j] = b[j];
                b[j] = 0;
            }
        }
        
        cout << a[n] << endl;
    }

    return 0;
}