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
?
解題思路：找兩個1~10進制之間的回文數(shù)字，當時看成找1個回文數(shù)字就可以通過，所以導致好久才AC，看題失誤！
1
#include <stdio.h>
2
#include <
string
.h>
3
4
char
A[
200
];
5
int
main()
6
{
7
int
num,r,i,n,j,t,k,ke,mark,len,last,flag;
8
scanf(
"
%d
"
,&
n);
9
while
(n--
){
10
scanf(
"
%d %d
"
,&last, &
k);
11
while
(last--
){
12
++
k;
13
14
flag=
0
;
15
for
(r=
2
;r<=
10
;r++
){
16
i=
0
;
17
num=
k;
18
mark=
1
;
19
20
while
(num>
0
){
21
t=num%
r;
22
A[i]= t+
'
0
'
;
23
++
i;
24
num/=
r;
25
}
26
len = i-
1
;
27
28
for
(i=
0
,j=len;i<=j;i++,j--
){
29
if
(A[i]!=
A[j])
30
mark=
0
;
31
}
32
33
if
(mark==
1
){
34
flag++
;
35
}
36
if
(flag==
2
){
37
printf(
"
%d\n
"
, k);
38
break
;
39
}
40
}
41
if
(flag!=
2
)
42
++
last;
43
}
44
}
45
return
0
;
46
}
?
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