1 second
256 megabytes
standard input
standard output
Teacher thinks that we make a lot of progress. Now we are even allowed to use decimal notation instead of counting sticks. After the test the teacher promised to show us a "very beautiful number". But the problem is, he's left his paper with the number in the
teachers' office.
The teacher remembers that the "very beautiful number" was strictly positive, didn't contain any leading zeroes, had the length of exactlyp decimal digits,
and if we move the last digit of the number to the beginning, it grows exactly x times. Besides, the teacher is sure that among all such numbers the "very
beautiful number" is minimal possible.
The teachers' office isn't near and the teacher isn't young. But we've passed the test and we deserved the right to see the "very beautiful number". Help to restore the justice, find the "very beautiful number" for us!
The single line contains integers p, x (1 ≤ p ≤ 106, 1 ≤ x ≤ 9).
If the teacher's made a mistake and such number doesn't exist, then print on a single line "Impossible" (without the quotes). Otherwise, print the "very beautiful
number" without leading zeroes.
6 5
142857
1 2
Impossible
6 4
102564
Sample 1: 142857·5 = 714285.
Sample 2: The number that consists of a single digit cannot stay what it is when multiplied by 2, thus, the answer to the test sample is "Impossible".
题意都能看明白，求一个数当它的最后一位放到第一位以后是x的倍数。
一道让人无语的题目，啥都不说了，枚举最后一位（1~9），倒着乘回去。
#include <stdio.h>
#include <string.h>
const int maxn = 1000010;
char s1[maxn],s2[maxn];
int main()
{
int p,x;
while(~scanf("%d %d",&p,&x))
{
int find = 0;
for(int i = 1; i <= 9; i++)
{
s1[0] = i+'0';
int c = 0;
for(int j = 0; j < p; j++)
{
int ans = (s1[j]-'0')*x+c;
s2[j] = ans%10+'0';
s1[j+1] = s2[j];
c = ans/10;
}
if(c > 0) continue;//位数大于p，不合法
if(s1[0] != s2[p-1]) continue;//不满足条件，不合法
if(s1[p-1] == '0') continue;
find = 1;
for(int j = p-1; j >= 0; j--)
printf("%c",s1[j]);
printf("\n");
break;
}
if(!find) printf("Impossible\n");
}
return 0;
}