c - 如何使一个变量等于C中其他变量的字符串

Question

这是完整的代码，挖掘变量是用户输入

#include <stdio.h>
#include <stdlib.h>

  int main()
{
    int num;
    int dig1=0;
    int dig2=0;
    int dig3=0;
    num=dig1,dig2,dig3;
    fflush(stdin);
    printf("Please enter the first digit of your three digit number:");
    scanf("%d", &dig1);
    fflush(stdin);
    printf("Please enter the second digit of your three digit number:");
    scanf("%d", &dig2);
    fflush(stdin);
    printf("Please enter the third digit of your three digit number:");
    scanf("%d", &dig3);

    if (num==(dig1*dig1*dig1)+(dig2*dig2*dig2)+(dig3*dig3*dig3))
    {
        printf("Your number is an Armstrong number!\n");
    }
    else
    {
        printf("Your number is not an Armstrong number!\n");
    }

    system("pause");
    return 0;
}

如何使变量“num”等于“dig1”、“dig2”和“dig3”的所有输入。就像如果 dig 1 是 2、dig 2 是 4、dig 3 是 6 一样，num 将是 246。请帮忙!

Answer 1

您的问题可以这样解决:

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

int main()
{
    int num;
    int dig[3];
    int i,j = 100;
    int result = 0;
    for(i = 0;i<3;i++){
        fflush(stdin);
        printf("Please enter the %d digit of your three digit number:",i+1);
        scanf("%d", &dig[i]);
    }

    for(i = 0;i<3;i++){
        result += pow(dig[i],3);
        num += dig[i] * j;
        j/=10;
        printf("%d * %d = %d\n",dig[i],j,num);
    }

    if (num == result)
    {
        printf("Your number is an Armstrong number!\n");
    }
    else
    {
        printf("Your number is not an Armstrong number!\n");
    }

    system("pause");
    return 0;
}