进制的转化

#include <stdio.h>

int main(){
    
    int num;
    printf("请输入数字:");
    scanf("%d",&num);
    
    int jinzhi = 16;
    
    //二进制
    int wei = 1;
    long long sum = 0;
    while(1){
        //取余 10100 
        // 00101
        // 1*0 + 10*0 + 100*1 + 1000*0 + 10000*1
        // 10100
        int yu = num % jinzhi;
        num = num / jinzhi; //10 5 
        
        //目前的总和：当前位数的值 + 之前的值 
        sum = sum + yu * wei;//0 
        wei = wei * 10 ;//1*10 100 1000 
        
        if(num < jinzhi){
            sum = sum + num * wei;
            break;
        } 
    } 
    printf("%d\n", sum);
    return 0;
}