Alice got a hold of an old calculator that can display n digits. She was bored enough to come up withthe following time waster.
She enters a number k then repeatedly squares it until the result overflows. When the resultoverflows, only the n most significant digits are displayed on the screen and an error flag appears. Alicecan clear the error and continue squaring the displayed number. She got bored by this soon enough,but wondered:
“Given n and k, what is the largest number I can get by wasting time in this manner?”
Input
The first line of the input contains an integer t (1 ≤ t ≤ 200), the number of test cases. Each test casecontains two integers n (1 ≤ n ≤ 9) and k (0 ≤ k < 10n) where n is the number of digits this calculatorcan display k is the starting number.
Output
For each test case, print the maximum number that Alice can get by repeatedly squaring the startingnumber as described.
Sample Input2
1 6
2 99
Sample Output
9
99
#include<stdio.h>
#include<string.h>
#include<algorithm>
#define INF 0x3f3f3f3f
#define N 10010
#define ll long long
using namespace std;
int a[110];
int next(int n,int k)
{
if(!k)
return 0;
ll k2=(ll)k*k;
int l=0;
while(k2>0)
{
a[l++]=k2%10;
k2/=10;
}
if(n>l)
n=l;
int ans=0;
for(int i=0;i<n;i++)
ans=ans*10+a[--l];
return ans;
}
int main()
{
int t,n,i,j,k;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&k);
int ans=k;
int k1=k;
int k2=k;
while(1)
{
k1=next(n,k1);//人1
k2=next(n,k2);//人2，第一步
if(k2>ans)
ans=k2;
k2=next(n,k2);//人2，第二步
if(k2>ans)
ans=k2;
if(k1==k2)//追上以后停止
break;
}
printf("%d\n",ans);
}
return 0;
}