Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Description
At the ruins of Wat Phra Si Sanphet (วดพระศรสรรเพชญ), one can find famous inscriptions that have only recently been decoded. Several numbers written using Thai numerals decorate these ruins.
A couple of years ago, the famous Peruvian researcher Renzo "el intrépido" Morales found out that most numbers found at the ruins are palindromic, that is, they represent the same number when read backwards. For instance, 171 is palindromic, whereas 17 is not.
Intrigued by the existence of non palidromic numbers as decorations in the ruins, Renzo found out that, while these numbers are not palindromic when represented in base 10 (which is the base used in the Thai numeral system), they are palindromic when represented in another base. For b > 0, the base-b representation of a number N is the sequence amam - 1... a1a0 so that 0 ≤ ai ≤ b - 1 for each 0 ≤ i ≤ m, am ≠ 0, and
ambm + am - 1bm - 1 + ... + a1b + a0 = N.For the previous example, the base-2 representation of 17 is 10001, which is palindromic.
To validate his discovery, Renzo wants you to write a program that takes a number represented in base 10 and checks in which bases from 2 to 16 such number has a palindromic representation.
Input
The first line has a single integer T, the number of test cases.
Each test case has a single line with an integer N written in base 10.
Output
For each test case, print in a single line a space-separated list of integers, from 2 to 16 and in increasing order, of the bases for which the representation of N is palindromic. If N does not have a palindromic representation for any of the bases from 2 to 16, print -1.
Limits
- 0 ≤ T ≤ 103
- 0 ≤ N < 231
Sample Input
2
17
2570
2 4 16
4 16
/*/
题意：
给你一个数，将这个数换成2~16进制的任何数，如果换成其他进制的数之后能够形成一个回文数【不懂去百度】就输出这个进制。 如果一个都不能形成回文数，输出 -1 。 模拟短除法，暴力解题。 AC代码：
/*/
#include"algorithm"
#include"iostream"
#include"cstring"
#include"cstdio"
#include"string"
#include"vector"
#include"queue"
#include"cmath"
using namespace std;
#define FK(x) cout<<"["<<x<<"]"<<endl
#define memset(x,y) memset(x,y,sizeof(x))
#define memcpy(x,y) memcpy(x,y,sizeof(x))
#define bigfor(x) for(int qq=1;qq<=x;qq++)
#define FIN freopen("input.txt","r",stdin)
#define FOUT freopen("output.txt","w+",stdout) int ans[20]; bool ok(int n,int i) {
memset(ans,0);
int erear=0;
while(n) {
ans[erear++]=n%i;
n/=i;
}
for(int j=0; j<erear/2; j++) {
if(ans[j]!=ans[erear-1-j])return 0;
}
return 1;
} int main() {
int T;
scanf("%d",&T);
bigfor(T) {
int n;
scanf("%d",&n);
int first=1;
int sign=1;
for(int i=2; i<17; i++) {
if(ok(n,i)) {
if(first) first=0;
else printf(" ");
printf("%d",i);
sign=0;
}
}
if(sign)puts("-1");
else puts("");
}
return 0;
}