C语言利用已知公式估算e的近似值

#include <math.h>
#include <stdio.h>

double calculateE(double precision) {
    double e = 1, temp = 1;
    int count = 1;
    for (int i = 1; i < 1000000; i++) {
        for (int j = 1; j <= i; j++) {
            temp *= j;
        }
        if (fabs(1.0 / temp) < precision) {
            break;
        }
        e += 1.0 / temp;
        count++;
        temp = 1;
    }
    return e;
}

int main() {

    double precision;
    printf("请输入精确度（例如10e-6）：");
    scanf("%le", &precision);
    double e=calculateE(precision);
    printf("e的近似值为%.6lf\n", e);
    return 0;
}