c语言编程阶乘计算求急

Question

1-2/3！+3/5！-4/7！+5/9！-....直到某一项绝对值小于10的-6次方为止

Answer 1

#include <stdio.h>
int main()
{
    double sum = 0, dx = 1, f = 1;

    long long n = 1, x = 1, i = 1;
    while ((dx >= 1e-6))
    {
        printf("%c%d/%d", (f > 0) ? '+' : '-', n, x);
        dx = 1.0 * n / x;
        sum += f * dx;

        n++;
        x *= ++i;
        x *= ++i;

        f = -f;
    }
    printf("%lf", sum);

    return 0;
}