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[200];
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=0;
            for(r=2;r<=10;r++){
                i=0;
                num=k;
                mark=1;
                
                while(num>0){
                    t=num%r;
                    A[i]= t+'0';    
                    ++i;
                    num/=r;
                }
                 len = i-1;

                for(i=0,j=len;i<=j;i++,j--){
                    if(A[i]!=A[j])
                        mark=0;
                }

                if(mark==1){
                    flag++;
                }
                if(flag==2){
                    printf("%d
", k);
                    break;                    
                }
            }
            if(flag!=2)
                ++last;
        }
    }
    return 0;
}