HDU-4394 Digital Square（DFS）

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4394

Digital Square

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

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

Sample Output

None115

数字n从个位开始的数字分别为n0,n1,n2……

模拟乘法运算，观察可得：

n0=e^2%10

n1=(2*d*e+e^2/10)%10注意进位

n2=(2*c*e+d^2+(2*d*e+e^2/10)/10)%10

=[2*c*e+(100*d^2+10*2*d*e+e^2)/100]%10

={2*c*e+[(de)^2]/100}%10

同理可得：n3,n4……

#include <cstdio>#include <cstring>#include <algorithm>//#define LOCALusing namespace std;const __int64 INF=1<<30;__int64 n,ans;int len,m;int num[10];void DFS(__int64 pre,__int64 temp,int pos)//①pre=10^pos,用于进位②temp表示当前已经确定的后pos位数字③pos表示第pos位数{    if(len==pos)    {        ans=min(ans,temp);        return;    }    for(int i=0;i<=9;++i)    {        if(num[pos]==(temp*temp/pre+m*i%10)%10)//如果第pos位数相符，则计算下一位            DFS(10*pre,temp+i*pre,pos+1);    }}int main(){#ifdef LOCALfreopen("input.txt","r",stdin);//freopen("output.txt","w",stdout);#endif    int T;    scanf("%d",&T);    while(T--)    {        scanf("%I64d",&n);        __int64 temp=n;        len=0;        while(temp)//计算n的每一位数字        {            num[len++]=temp%10;            temp/=10;        }        ans=INF;//给答案附最大值        for(int i=0;i<=9;++i)            if(num[0]==i*i%10)            {                m=2*i;                DFS(10,i,1);            }        if(ans==INF)            printf("None\n");        else            printf("%d\n",ans);    }    return 0;}

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