【HDU 1012】【水题】u Calculate e

#include "stdio.h"
int main(int argc, char const *argv[])
{
    int n=0,i=1,m=0,ij=1,i2=0;
    double cij;
    double answer=2.5;
    printf("n e\n- -----------\n0 1\n1 2\n");
    while ( m < 8)
    {
        ++m;
        while (i2 <= i)
        {
            ++i2;
            ij=ij*i2;  
        }
        ++i;
        cij=(1.0/ij);
        if (cij==0.5)
            printf("%d %.1f\n",2,2.5);
        else 
        {
            answer=answer+cij;
            printf("%d %.9lf\n",i,answer);
        }
    }
    return 0;
}

水题，注意下精度什么的就好，会看懂数学符号应该就可以。