蓝桥杯历届试题-回文数字 （C语言代码）（最笨但也最简单的方法）

解题思路:
       本题条件不多，完全可以通过简单的列举法
注意事项:

参考代码:

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

int main()
{
     int n;
     int ch[6];
     unsigned int i;
     int j,sum;
     int k=0;
     int flag;//标志变量
     printf("Input n:(10< n< 100)\n");
     scanf("%d",&n);
     while(n<=10||n>=100)
     {
         scanf("%d",&n);
     }
     for(i=10000;i<=99999;i++)
     {
         flag=0;
         sum=0;
         ch[4]=i%10;
         ch[0]=i/10000;
         ch[1]=(i-ch[0]*10000)/1000;
         ch[2]=(i-ch[0]*10000-ch[1]*1000)/100;
         ch[3]=(i-ch[0]*10000-ch[1]*1000-ch[2]*100)/10;
         for(j=0;j<3;j++)
         {
             if(ch[j]!=ch[4-j])
                flag=1;
         }
         if(flag==1)
            continue;
         for(j=0;j<5;j++)
         {
             sum=sum+ch[j];
         }
         if(sum==n)
         {
             k++;
             printf("%d\n",i);
         }
     }
     for(i=100000;i<=999999;i++)
     {
         flag=0;
         sum=0;
         ch[5]=i%10;
         ch[0]=i/100000;
         ch[1]=(i-ch[0]*100000)/10000;
         ch[2]=(i-ch[0]*100000-ch[1]*10000)/1000;
         ch[3]=(i-ch[0]*100000-ch[1]*10000-ch[2]*1000)/100;
         ch[4]=(i-ch[0]*100000-ch[1]*10000-ch[2]*1000-ch[3]*100)/10;
         for(j=0;j<3;j++)
         {
             if(ch[j]!=ch[5-j])
                flag=1;
         }
         if(flag==1)
            continue;
         for(j=0;j<6;j++)
         {
             sum=sum+ch[j];
         }
         if(sum==n)
         {
             k++;
             printf("%d\n",i);
         }
     }
     if(k==0)
     {
         printf("-1\n");
     }
     return 0;
}