Question

我试图用C编写一个程序，该程序找到两个数字的最小除数。我想我缺少一些基本的知识来使它正常工作，但是我不确定是什么。

#include <stdio.h>

int main(void){
    int divisor = 2;
    int dividend = 0;
    int answer;
    printf("%s", "Please enter a number we will use as a dividend.");
    scanf("%d", &dividend);
    while (divisor != 0) {
        answer = dividend / divisor;
        if (answer == 0)
        else divisor++;
    }
    printf("The lowest positive divisor is: ");
}

Answer 1

您的代码很好，这只是您的中断条件。应该是这样的：

while (divisor != 0){
    if(dividend % divisor == 0)
        break;
    else divisor++;
}

Answer 2

问题陈述中肯定会有一些困惑：两个数字的最小除数是1，如果数字没有共同的主除数，则可能不会有更大的除数。

要找到2个数字的最大公约数，可以使用Euclid's algorithm：

#include <stdio.h>

int main(void) {
    int n1, n2;
    printf("%s", "Please enter two numbers: ");
    if (scanf("%d", &n1, &n2) == 2) {
        int a = n1, b = n2;
        while (b != 0) {
            int c = a % b;
            a = b;
            b = c;
        }
        printf("gcd(%d, %d) = %d\n", n1, n2, a);
    }
    return 0;
}

要找到大于1的最低除数，必须在循环中测试两个数字：

#include <stdio.h>

int main(void) {
    int n1, n2;
    printf("%s", "Please enter two numbers: ");
    if (scanf("%d", &n1, &n2) == 2) {
        int a = 2, res = 1;
        for (;; a++) {
            if (a > n1 / a)  // no larger divisor of n1
                break;
            if (n1 % a != 0) // not a divisor of n1
                continue;
            if (a > n2 / a)  // no larger divisor of n2
                break;
            if (n2 % a != 0) // not a divisor of n2
                continue;
            res = a;         // found the lowest common divisor > 1
            break;
        }
        printf("lowestcd(%d, %d) = %d\n", n1, n2, a);
    }
    return 0;
}

用C程序查找两个数字的最小除数

2 个答案: