hdu4394Digital Square(优先队列+广搜+__int64)

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%10^x=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.

 

 

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

2012 Multi-University Training Contest 10

 题意：给出n，求出最小的m，满足m^2  % 10^k = n，其中k=0,1,2

思路： 广搜，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，可见可以一位一位的判断。

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。

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <queue>
using namespace std;
__int64 n;
struct node
{
    int num;   //判断到第几位
    __int64 w;   //当前判断的数字
};
node node1,node2;
 bool operator<(node n1,node n2)  //优先队列，必须输出最小的数
    {
        return n1.w>n2.w;   //从小到大
    }

void bfs()
{
     priority_queue <node> q;
     __int64 t;
     while(!q.empty())
     q.pop();
     node1.num=0;
     node1.w=0;
     q.push(node1);
     while(!q.empty())
     {
         node1=q.top();//优先队列用q.top(),队列用q.front()
         q.pop();
         t=(__int64)pow(10.0,node1.num);
         if(node1.w*node1.w%t==n)
         {
             printf("%I64d\n",node1.w);
             return ;
         }
         for(int i=0;i<10;i++)
         {
             node2.num=node1.num+1;//向右增加一位
             node2.w=node1.w+t*i;  //得到增加一位的数字
             if(node2.w*node2.w%(t*10)==n%(10*t))
             q.push(node2);
         }
     }
     printf("None\n");
}
int main()
{
    int m;
    scanf("%d",&m);
    while(m--)
    {
        scanf("%I64d",&n);
        if(n%10==2||n%10==3||n%10==7||n%10==8)
        printf("None\n");
        else
        bfs();
    }
}

　　

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