从 5 个整数中计算偶数的 C 程序

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我正在尝试编写一个 C 程序，其中用户输入五个不同的整数，并根据这五个整数的输入确定偶数整数的数量。这是我当前的代码:

#include <stdio.h> 

int main()
{
    int n1, n2, n3, n4, n5, sum;

//user enters 5 integers

    printf("Enter five different positive integers: \n");

//program scans for user input

    scanf("%d %d %d %d %d", &n1, &n2, &n3, &n4, &n5);

//if statement to determine what integers are even

    if(((n1,n2,n3,n4,n5)%2)==0)

//sum of even integers

        sum = n1 + n2 + n3 + n4 + n5;

//program prints sum of even integers

        printf("There are %d even integers in the input. \n", sum); 

//program prints if there are no even integers for the inputs

    else

        printf("There are no even integers in the input. \n");

    return(0);
}

有什么想法吗？

Answer 1

您的目标没有根据需要明确表述:

• 您想对所有偶数求和并忽略输入的奇数吗？

• 是否希望所有整数都是偶数，如果没有则拒绝输入？

无论哪种方式，您的程序都会因多种原因而失败:

• if(((n1,n2,n3,n4,n5)%2)==0) 没有任何用处:它只检查最后一个整数是否为偶数。您可以检查所有整数是否都是偶数

   if ((n1 | n2 | n3 | n4 | n5) % 2) == 0)
  

• 您没有使用大括号对 if 正文中的指令进行分组。与 Python 不同，缩进在 C 中不起作用，您必须在多个指令周围使用大括号({ 和 })在 if 之后形成一个 block ， else、while、for 等

这是代码的修改版本，忽略奇数:

#include <stdio.h> 

int main(void) {
    int n1, n2, n3, n4, n5, sum, count;

    // user enters 5 integers
    printf("Enter five different positive integers:\n");

    // program scans for user input

    if (scanf("%d %d %d %d %d", &n1, &n2, &n3, &n4, &n5) != 5) {
        printf("Invalid input\n");
        return 1;
    }

    // for each integer, add it if it is even

    count = 0;
    sum = 0;

    if (n1 % 2 == 0) {
        sum += n1;
        count++;
    }
    if (n2 % 2 == 0) {
        sum += n2;
        count++;
    }
    if (n3 % 2 == 0) {
        sum += n3;
        count++;
    }
    if (n4 % 2 == 0) {
        sum += n4;
        count++;
    }
    if (n5 % 2 == 0) {
        sum += n5;
        count++;
    }

    if (count > 0) {
        printf("There are %d even integers in the input, their sum is %d.\n",
               count, sum); 
    } else {
        //program prints if there are no even integers for the inputs
        printf("There are no even integers in the input.\n");
    }
    return 0;
}

使用一些更高级的 C 语言知识，您可以将代码简化为:

#include <stdio.h> 

int main(void) {
    int n1, n2, n3, n4, n5, sum, count;

    printf("Enter five different positive integers:\n");
    if (scanf("%d %d %d %d %d", &n1, &n2, &n3, &n4, &n5) != 5) {
        printf("Invalid input\n");
        return 1;
    }

    // use the low order bit to test oddness
    count = 5 - ((n1 & 1) + (n2 & 1) + (n3 & 1) + (n4 & 1) + (n5 & 1));
    sum = n1 * !(n1 & 1) + n2 * !(n2 & 1) + n3 * !(n3 & 1) +
          n4 * !(n4 & 1) + n4 * !(n4 & 1);

    if (count > 0) {
        printf("There are %d even integers in the input, their sum is %d.\n",
               count, sum); 
    } else {
        printf("There are no even integers in the input.\n");
    }
    return 0;
}

但它实际上更复杂，可读性更差，而且效率也没有被证明更高。

真正的改进是使用循环:

#include <stdio.h> 

int main(void) {
    int i, n, sum = 0, count = 0;

    printf("Enter five different positive integers:\n");
    for (i = 0; i < 5; i++) {
        if (scanf("%d, &n) != 1) {
            printf("Invalid input\n");
            return 1;
        }
        if (n % 2 == 0) {
            sum += n;
            count++;
        }
    }

    if (count > 0) {
        printf("There are %d even integers in the input, their sum is %d.\n",
               count, sum); 
    } else {
        printf("There are no even integers in the input.\n");
    }
    return 0;
}