HDOJ 4394 —— BFS

Digital Square

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 830    Accepted Submission(s): 349

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

332125

 

Sample Output

None115

 

Source

2012 Multi-University Training Contest 10

 

Recommend

zhuyuanchen520

 

题意是给你一个N，让你输出最小的M使得   M^2 % 10^x = N

我们从最低位向高位去搜，如果  M^2 % 10^y = N % 10^y  则将之加入队列。

    #include <cstdio>    #include <cmath>    #include <algorithm>    #include <iostream>    #include <cstring>    #include <map>    #include <string>    #include <stack>    #include <cctype>    #include <vector>    #include <queue>    #include <set>    #include <utility>    using namespace std;    //#define Online_Judge    #define outstars cout << "***********************" << endl;    #define clr(a,b) memset(a,b,sizeof(a))    #define lson l , mid  , rt << 1    #define rson mid + 1 , r , rt << 1 | 1    //#define mid ((l + r) >> 1)    #define mk make_pair    #define FOR(i , x , n) for(int i = (x) ; i < (n) ; i++)    #define FORR(i , x , n) for(int i = (x) ; i <= (n) ; i++)    #define REP(i , x , n) for(int i = (x) ; i > (n) ; i--)    #define REPP(i ,x , n) for(int i = (x) ; i >= (n) ; i--)    const int MAXN = 100000 + 50;    const long long LLMAX = 0x7fffffffffffffffLL;    const long long LLMIN = 0x8000000000000000LL;    const int INF = 0x3f3f3f3f;    const int IMIN = 0x80000000;    const double E = 2.718281828;    #define eps 1e-8    #define DEBUG 1    #define mod 100000007    typedef long long LL;    const double PI = acos(-1.0);    typedef double D;    typedef pair<int , int> pi;    //    #pragma comment(linker, "/STACK:102400000,102400000")__int64 a[10050];    struct Node    {        __int64 m , x;        Node()        {            m = 0;            x = 0;        }    };    __int64 ten[10];    void init()    {        __int64 num = 1;        FOR(i , 0 , 10)        {            ten[i]= num;            num *= 10;        }    }    __int64 getlength(__int64 x)    {        if(x == 0)return 1;        int num = 0;        while(x > 0)        {            x /= 10;            num++;        }        return num;    }    int main()    {        int t;        __int64 n;        cin >> t;        init();        while(t--)        {            scanf("%I64d" , &n);            int len = getlength(n);            queue <Node> q;            q.push(Node());//            outstars            while(!q.empty())            {                Node u = q.front();//                q.pop();//                outstars                if(u.x == len)                {                    break;                }//                outstars                FOR(i , 0 , 10)                {                    Node v ;                    v.m = u.m + ten[u.x] * i;                    v.x = u.x + 1;//                    outstars                    if((v.m * v.m) % ten[v.x] == n % ten[v.x])                    {//                        outstars                        q.push(v);//                        outstars                    }                }                q.pop();//                outstars            }//            outstars            __int64 ans = INF;            while(!q.empty())            {                Node u = q.front();                q.pop();                if(ans > u.m)ans = u.m;            }            if(ans == INF)printf("None\n");            else printf("%I64d\n" , ans);        }        return 0;    }