DFS+数学：Digital Square

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本文介绍了一个算法挑战，旨在寻找满足特定条件的最小非负整数M，使得M的平方数的尾数等于给定的整数N。通过递归深度优先搜索的方法，实现了对所有可能的尾数进行遍历，最终找到符合条件的最小整数。

Digital Square

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 743 Accepted Submission(s): 139

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

None
11
5

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>

using namespace std;

int n;
long long ans = 10000000000LL;

long long dfs(long long shi, long long now){
    if(shi > n)
        return 0;
    int sheng = n % shi;
    for(long long i = 0; i <= 9; i ++){
        long long a = i * shi + now;
        long long aa = a * a % (shi * 10);
        long long nn = n % (shi * 10);
        if(aa == nn){
            //printf("a=%I64d aa=%I64d shi=%I64d\n", a, aa, shi);
            if(aa == n){
                ans = min(ans, a);
            }
            else
                dfs(shi * 10, a);
        }
    }
    return 0;
}

int main(){
    int t;
    scanf("%d", &t);
    while(t --){
            ans = 10000000000LL;
        scanf("%d", &n);
        if(n == 0){
            printf("0\n");
            continue;
        }
        if(n == 1000000000){
            printf("None\n");
            continue;
        }
        dfs(1, 0);
        if(ans == 10000000000LL)
            printf("None\n");
        else
            printf("%I64d\n", ans);
    }
    return 0;
}