Digital Square
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1827 Accepted Submission(s):
714
Problem 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
Source
Recommend
zhuyuanchen520 | We have carefully selected several
similar problems for you: 4390 4398 4397 4396 4395
搜索题，首先证明出N位后缀只与M的后N位有关。比如三位数100a+10b+c平方后展开为 10000a^2+2000ab+100b^2+200ac+20bc+c^2很显然，平方后的最后一位只与c有关最后两位只与bc有关，最后三位abc都有关。
那我们只需要BFS一下，不断地找满足最后指定位数的数，1位，2位，……直到找到第一个满足条件的。
题意：给出n，求出最小的m，满足m^2 % 10^k = n，其中k=0,1,2
附上代码：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <queue> 6 using namespace std; 7 __int64 n; 8 struct node 9 { 10 __int64 a; 11 int x; 12 friend bool operator < (node n1,node n2) //优先队列，必须输出最小的数 13 { 14 return n1.a>n2.a; 15 } 16 } s1,s2; 17 18 void BFS() 19 { 20 priority_queue <node> q; 21 while(!q.empty()) 22 q.pop(); 23 __int64 t; 24 s1.a=0; 25 s1.x=0; 26 q.push(s1); 27 while(!q.empty()) 28 { 29 s1=q.top(); 30 q.pop(); 31 t=(__int64)pow(10.0,s1.x); //从低位向高位搜 32 if(s1.a*s1.a%t==n) //找到了就直接输出 33 { 34 printf("%I64d ",s1.a); 35 return; 36 } 37 for(int i=0; i<10; i++) 38 { 39 s2.x=s1.x+1; //每次都向前进一位 40 s2.a=s1.a+i*t; 41 if(s2.a*s2.a%(t*10)==n%(t*10)) //一位一位的匹配，匹配成功后，则放入队列 42 q.push(s2); 43 } 44 } 45 printf("None "); 46 } 47 int main() 48 { 49 int T,m,i,j; 50 scanf("%d",&T); 51 while(T--) 52 { 53 scanf("%d",&n); 54 if(n%10==2||n%10==3||n%10==7||n%10==8) //根据乘法的规律，任何一个数平方后的个位数不可能是这些数 55 { 56 printf("None "); 57 continue; 58 } 59 BFS(); 60 } 61 return 0; 62 }