"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!" 

# include <stdio.h>

int main()

{

    int C1[], C2[], n;

    while(scanf("%d", &n) != EOF)

    {

        for(int i = ; i <= n; i++)//初始化 第一个表达式 目前所有指数项的系数都为1

        {

            C1[i] = ;

            C2[i] = ;

        }

        for(int i = ; i <= n; i++)//第2至第n个表达式

        {

            for(int j = ; j <= n; j++)//C1为前i-1个表达式累乘后各个指数项的系数

            {

                for(int k = ; j + k <= n; k += i)//k为第i个表达式每个项的指数 第一项为1【即x^(i * 0)】（指数k=0），第二项为x^(i * 1)（指数为k=i）， 第三项为x^(i * 2)... 所以k的步长为i

                {

                    C2[j + k] += C1[j];//(ax^j)*(x^k) = ax^(j+k) -> C2[j+k] += a  【第i个表达式每一项的系数都为1； a为C1[j]的值（x^j的系数）； C2为乘上第i个表达式后各指数项的系数】

                }

            }

            for(int j = ; j <= n; j++)//刷新当前累乘结果各指数项的系数

            {

                C1[j] = C2[j];

                C2[j] = ;

            }

        }

        printf("%d\n",C1[n]);

    }

    return ;

}