Description
Given an integer N,you should come up with the minimum nonnegative integer M.M meets the follow condition: M2%10x=N (x=0,1,2,3....)
Input
The first line has an integer T( T< = 1000), the number of test cases.
For each case, each line contains one integer N(0<= N <=109), indicating the given number.
For each case, each line contains one integer N(0<= N <=109), indicating the given number.
Output
For each case output the answer if it exists, otherwise print “None”.
Sample Input
3
3
21
25
Sample Output
None
11
5
1 #include <iostream> 2 #include <stdio.h> 3 #include <set> 4 #include <string.h> 5 #include <algorithm> 6 #include <queue> 7 using namespace std; 8 #define ll long long 9 ll n,nu,ans; 10 void init() 11 { 12 ll nn=n; 13 nu=1; 14 while(nn) 15 { 16 nn/=10; 17 nu*=10; 18 } 19 } 20 bool fun() 21 { 22 int i,ok=0; 23 ans=n; 24 ll now,x,y,noww; 25 queue<pair<ll,ll> >q; 26 while(!q.empty()) 27 q.pop(); 28 q.push(make_pair(0,1)); 29 while(!q.empty()) 30 { 31 x=q.front().first; 32 now=q.front().second; 33 noww=now*10; 34 q.pop(); 35 for(i=0;i<10;i++) 36 { 37 y=x+now*i; 38 if(y*y%nu==n){ans=min(ans,y);ok=1;} 39 else 40 if(y*y%noww==n%noww)q.push(make_pair(y,noww)); 41 } 42 } 43 if(ok)return 1; 44 return 0; 45 } 46 int main() 47 { 48 int t; 49 scanf("%d",&t); 50 while(t--) 51 { 52 cin>>n; 53 if(n==0) 54 { 55 cout<<10<<endl; 56 continue; 57 } 58 init(); 59 if(fun()) 60 { 61 cout<<ans<<endl; 62 } 63 else printf("None "); 64 } 65 }