c语言-分别用while语句、do-while语句、for语句编写程序,计算m=1+1/1!+1/2!+...+1/n!

以下是三种C语言计算m的方法：
1. while循环实现

#include <stdio.h>
int main()
{
    double sum = 0, a = 1;
    int i = 1, n = 20;
    while(i <= n)
    {
        a *= i;
        sum += 1.0 / a;
        i++;
    }
    printf("%lf", sum);
    return 0;
}

2. do-while循环实现

#include <stdio.h>
int main()
{
    double sum = 0, a = 1;
    int i = 1, n = 20;
    do
    {
        a *= i;
        sum += 1.0 / a;
        i++;
    }while(i <= n);
    printf("%lf", sum);
    return 0;
}

3. for循环实现

#include <stdio.h>
int main()
{
    double sum = 0, a = 1;
    int i, n = 20;
    for(i = 1; i <= n; i++)
    {
        a *= i;
        sum += 1.0 / a;
    }
    printf("%lf", sum);
    return 0;
}

相关问题

c语言-分别用while语句、do-while语句、for语句编写程序,计算e = 1+1/1!+1/2!+...+1/n!

以下是三种C语言计算e的程序，分别使用while语句、do-while语句、for语句实现：
1.使用while语句实现：

#include <stdio.h>

int main() {
    int n = 20, i = 1;
    double sum = 1, a = 1;
    while (i <= n {
        a *= i;
        sum += 1.0 / a;
        i++;
    }
    printf("e = %lf\n", sum);
    return 0;
}

2.使用do-while语句实现：

#include <stdio.h>

int main() {
    int n = 20, i = 1;
    double sum = 1, a = 1;
    do {
        a *= i;
        sum += 1.0 / a;
        i++;
    } while (i <= n);
    printf("e = %lf\n", sum);
    return 0;
}

3.使用for语句实现：

#include <stdio.h>

int main() {
    int n = 20;
    double sum = 1, a = 1;
    for (int i = 1; i <= n; i++) {
        a *= i;
        sum += 1.0 / a;
    }
    printf("e = %lf\n", sum);
    return 0;
}

c语言-分别用while语句、do-while语句、for语句编写程序,计算e≈1+1/1!+1/2!+...+1/n!

以下是三种C语言计算e的程序，分别使用while语句、do-while语句、for语句实现：
1.使用while语句实现：

#include <stdio.h>

int main() {
    int n = 1;
    double sum = 1, a = 1;
    while (n <= 20) {
        a *= n;
        sum += 1.0 / a;
        n++;
    }
    printf("e = %lf\n", sum);
    return 0;
}

2.使用do-while语句实现：

#include <stdio.h>

int main() {
    int n = 1;
    double sum = 1, a = 1;
    do {
        a *= n;
        sum += 1.0 / a;
        n++;
    } while (n <= 20);
    printf("e = %lf\n", sum);
    return 0;
}

3.使用for语句实现：

#include <stdio.h>

int main() {
    int n;
    double sum = 1, a = 1;
    for (n = 1; n <= 20; n++) {
        a *= n;
        sum += 1.0 / a;
    }
    printf("e = %lf\n", sum);
    return 0;
}