C 语言计算10个数字之和并跳过负数

编写一个C程序，使用for循环计算10个数字之和并跳过负数。在此C示例中，for循环从1迭代到10，if语句检查数字是否小于零。如果为真，continue语句将跳过该数字以执行加法。

#include <stdio.h>

int main()
{   
    int num, sum = 0;
    
    printf("Please Enter the 10 Numbers\n");
    for(int i = 1; i <= 10; i++)
    {
        printf("Number %d = ", i);
        scanf("%d", &num);

        if(num < 0)
        {
            continue;
        }
        sum = sum + num;
    }

    printf("\nSum by Skipping Negative Numbers = %d\n", sum); 
}

此C程序从用户输入读取十个数字，并使用while循环计算10个数字之和，同时跳过负数。

#include <stdio.h>

int main()
{   
    int i, num, sum = 0;
    
    printf("Please Enter the 10 Numbers\n");
    i = 1; 
    while(i <= 10)
    {
        printf("Number %d = ", i);
        scanf("%d", &num);

        if(num < 0)
        {
            i++;
            continue;
        }
        sum = sum + num;
        i++;
    }

    printf("\nSum by Skipping Negative Numbers = %d\n", sum); 
}

Please Enter the 10 Numbers
Number 1 = 12
Number 2 = 23
Number 3 = -99
Number 4 = 14
Number 5 = -3
Number 6 = 77
Number 7 = 88
Number 8 = -98
Number 9 = -14
Number 10 = 109

Sum by Skipping Negative Numbers = 323