HDU 5373 The shortest problem

分析：题目问最后得到的数能不能被11整除，11的倍数的特点是奇数位和偶数位的差能被11整除，知道这就好求了。

# include <stdio.h>
# include <string.h>
  char s[10];
  int main()
  {
      int i,n,t,len,cas=1,sum,tem,k,s1,s2,x1,x2;
      while(scanf("%d%d",&n,&t))
      {
          if(n==-1&&t==-1)
            break;
          sprintf(s,"%d",n);
          len=strlen(s);
          s1=s2=0;
          for(i=sum=0;i<len;i++)
          {
              sum+=s[i]-'0';
              if(i%2==0)
                s1+=s[i]-'0';
              else
                s2+=s[i]-'0';
          }
          while(t--)
          {
              tem=sum;
              k=x1=x2=0;
              while(tem)
              {
                  if(k%2==0)
                    x1+=tem%10;
                  else
                    x2+=tem%10;
                  sum+=tem%10;
                  tem/=10;
                  k++;
              }
              if(len%2==0)
              {
                  if(k%2==1)
                    s1+=x1,s2+=x2;
                  else
                    s1+=x2,s2+=x1;
              }
              else
              {
                  if(k%2==1)
                    s1+=x2,s2+=x1;
                  else
                    s1+=x1,s2+=x2;
              }
              len+=k;
          }
          printf("Case #%d: ",cas++);
          if((s1-s2)%11==0)
            printf("Yes\n");
          else
            printf("No\n");
      }
      return 0;
  }