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C/C++ code

//刚才看到一道题//自己写了代码结果和答案不一样//希望大家帮我看下麻烦了//题目是这样的:  个位数为6且能被3整除的五位数共有多少？//我写的#include <iostream>using namespace std;int main(){    int      result = 0, one = 0;    for (long i = 10000; i != 99999; i++)    {        one = i % 10;//先求这个数的个位数        if ( (one == 6) && (one % 3) == 0)//判断满足条件的        {            result++;        }    }    cout << result << endl;    return  0;}//================================================/*别人给出的答案根据题意可知，满足条件的五位数的选择范围是10006、10016。。。99996。可设基础数i=1000，通过计算i*10+6即可得到欲选的数(i的变化范围是1000~999)，再判断该数能否被3整除。*/#include <iostream>using namespace std;int main(){    int      result = 0;    for (long i = 1000; i != 9999; i++)    {        if ( (i * 10 + 6) % 3 == 0)        {            result++;        }    }    cout << result << endl;    return  0;}

C/C++ code

#include <iostream>using namespace std;int main(){    int      result = 0, one = 0;    for (long i = 10000; i != 99999; i++)    {        one = i % 10;//先求这个数的个位数        if ( (one == 6) && (one % 3) == 0)//判断满足条件的  // [color=#FF0000]这里应该是 i % 3吧[/color]        {            result++;        }    }    cout << result << endl;    return  0;}
[解决办法]
[code=C/C++][/code]
#include <iostream>

using namespace std;

int main()
{
   int      result = 0;
int      a=0,  b=0,   c=0,   d=0;
   for (long i = 10000; i != 99999; i++)
   {
       a = i%10000;                              /*这样求出个位数8*/
b = (i-a*10000)%1000;
c = (i-a*10000-b*1000)%100;
d = (i-a*10000-b*1000-c*100)%10;    
      
if ( (d == 6) && (i % 3) == 0)//判断满足条件的
       {
cout<<i<<endl;
           result++;
       }

   }

   cout << result << endl;

   return  0;
}
[code=C/C++][/code]

lz这样看看呢！
[解决办法]

int main()
{
int result,one;
result=0;
one=0;
for(long i=10000;i<=99999;i++)
{one=i%10;
if((one==6)&&(i%3)==0)
{
result++;
}
}
cout<<result<<endl;
  return 0;
}结果是3000
但是我个人觉得还是下面的代码比较清晰
int main()
{
   int      result = 0;

   for (long i = 1000; i != 9999; i++)
   {
       if ( (i * 10 + 6) % 3 == 0)
       {
           result++;
       }

   }

   cout << result << endl;

   return  0;
}