SZU:B54 Dual Palindromes

Judge Info

• Memory Limit: 32768KB
• Case Time Limit: 10000MS
• Time Limit: 10000MS
• Judger: Number Only Judger

Description

A number that reads the same from right to left as when read from left to right is called a palindrome. The number 12321 is a palindrome; the number 77778 is not. Of course, palindromes have neither leading nor trailing zeroes, so 0220 is not a palindrome.

The number 21 (base 10) is not palindrome in base 10, but the number 21 (base 10) is, in fact, a palindrome in base 2 (10101).

Write a program that reads two numbers (expressed in base 10):

• N (1 <= N <= 15)
• S (0 < S < 10000)

and then finds and prints (in base 10) the first N numbers strictly greater than S that are palindromic when written in two or more number bases (2 <= base <= 10). Solutions to this problem do not require manipulating integers larger than the standard 32 bits.

Input

The first line of input contains , the number of test cases.

For each test case, there is a single line with space separated integers N and S.

Output

For each test case output N lines, each with a base 10 number that is palindromic when expressed in at least two of the bases 2..10. The numbers should be listed in order from smallest to largest.

Sample Input

2
3 25
1 25

Sample Output

26
27
28
26

解题思路：找两个1~10进制之间的回文数字，当时看成找1个回文数字就可以通过，所以导致好久才AC，看题失误！

 #include <stdio.h>
 #include <string.h>
 
 char A[];
 int main()
 {
     int num,r,i,n,j,t,k,ke,mark,len,last,flag;
     scanf("%d",&n);
     while(n--){
         scanf("%d %d",&last, &k);
         while(last--){
             ++k;
 
             flag=;
             for(r=;r<=;r++){
                 i=;
                 num=k;
                 mark=;
 
                 while(num>){
                     t=num%r;
                     A[i]= t+'';
                     ++i;
                     num/=r;
                 }
                  len = i-;
 
                 for(i=,j=len;i<=j;i++,j--){
                     if(A[i]!=A[j])
                         mark=;
                 }
 
                 if(mark==){
                     flag++;
                 }
                 if(flag==){
                     printf("%d\n", k);
                     break;
                 }
             }
             if(flag!=)
                 ++last;
         }
     }
     return ;
 }