计算所有三位数中能被3整除且十位数字不是3的数之和并输出。很急用，请高手帮忙解答，谢谢。

int i, k, sum = 0;
for (i = 100; i < 1000; i ++)
{
    if (!(i%3))
    {
        k = i;
        k /= 10;
        if (k%10 != 3)
        {
            sum += i;        
        }
    }
}
printf("sum = %d\n", sum);

谢谢版主

#include<stdio.h>
#include<stdlib.h>
void main()
{
    int number,SUM=0;
    char str_number[3];
    for(number=100;number<=999;number++)
    {
        itoa(number,str_number,10);//将整数转换为字符串//
        {
            int k=number,sum=0;
            while(k>0)
            {
                sum+=k%10;
                k=k/10;
            }
            if((sum%3==0)&&(str_number[1]!='3'))//判断题中的条件//
                SUM+=number;
         }
    }
    printf("和是:%d\n",SUM);
}

#include<stdio.h>

void main()
{  
    int a,b,c;//百位，十位，个位
    int num;
    int sum=0;
    for(num=100;num<=999;num++)
    {  
        if(num%3==0){
            a=num/100;
            b=(num%100)/10;
            c=(num%100)%10;
            if(b!=3){
        printf("%5d",num);
        sum+=num;
            }
        }
        
    }
    printf("\n这些数的和为%d",sum);
}

#include "stdio.h"
main()
{
    int i,a2;
    float sum=0;
    for(i=100;i<1000;i++)
    {
        if(i%3==0)
        {
            
            a2=i%100/10;//十位数
        
            if(a2!=3)
            {
                printf("%4d",i);
                printf("\n");
            
                sum+=i;
            }
        }
    }
    printf("%lf  \n",sum);

}

#include <stdio.h>
void main()
{
    int i,sum=0,a[300];
    int k=0;
    for(i=100;i<1000;i++)
    {
        if(i%3==0 && (i/10)%10!=3)
        {
            a[k]=i;
            k++;
            sum+=i;
        }
    }
    for(i=0;i<k;i++)         //输出相应三位数
        printf("%4d",a[i]);
    printf("\nsum=%d\n",sum);      
}