HUST-OJ-1669

L - Palindromes
Time Limit:1000MS     Memory Limit:524288KB     64bit IO Format:%lld & %llu
Submit Status Practice HUST
1669

Description

Palindromes are numbers that read the same forwards as backwards. Like 12321.

Given a number base B (2 <= B <= 20), calculate all the number N( N no large than 300) such that N^2 is palindromic when expressed in base B. 

Pay attention, 'A', 'B', and so on represent the digits 10, 11, and so on.

Print both the number and its square in base B.

 

Input

A single line with B.

Output

Lines with two integers represented in base B. The first integer is the number whose square is palindromic; the second integer is the square itself.

Sample Input

10

Sample Output

1 1
2 4
3 9
11 121
22 484
26 676
101 10201
111 12321
121 14641
202 40804
212 44944

264 69696

题目不难 大意就是找出B进制的回文数，注意x和x^2都要回文数表示
#include<stdio.h>

int main()
{
int a[20],c[20];
int b,i,j,k,q,p;
scanf("%d",&b);
for(i=1;i<300;i++)
{
int x,y,flag=1;
x = i*i;
for(j=0;x!=0;j++)
{
a[j] = x%b;
x = x/b;
}
j--;
for(k=0;j>k;j--,k++)
if(a[j]!=a[k])
{
flag = 0;
break;
}
if(flag==1)
{
q = i;
for(j=0;q!=0;j++)
{
p = q%b;
c[j] = p;
q = q/b;
}
j--;
while(j>=0)
{
if(c[j]>9)
printf("%c",c[j]+55);
else
printf("%c",c[j]+48);
j--;
}
printf(" ");
x = i*i;
for(j=0;x!=0;j++){
y = x%b;
if(y>9)
printf("%c",y+55);
else
printf("%c",y+48);
x = x/b;
}
printf("\n");
}

}

}