未补完
E. Square
算法：二分
做法：我们依据x来枚举k，得到 \(x * 10^k\) ，用二分在[0, 1e9]查找mid的平方值，且这个值是第一个大于等于 \(x * 10^k\) 的值。得到这个值后我们再判断这个值在除 \(10^k\) 后是否与 \(x\) 相等即可。
code
#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <map>
#include <queue>
#include <stack>
#include <deque>
#include <cmath>
#include <string>
#include <set>
#define fir first
#define sec second
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<double, int> PDI;
typedef pair<double, double> PDD;
int dx[4] = { -1, 1, 0, 0 }, dy[4] = { 0, 0, -1, 1 };
int getlen(LL x)
{
int cnt = 0;
while(x)x /= 10, cnt++;
return cnt;
}
LL sl(LL x)
{
LL l = 0, r = 1e9;
while(l < r)
{
LL mid = (l + r) / 2;
if(mid * mid >= x)r = mid;
else l = mid + 1;
}
return l;
}
LL sr(LL x)
{
LL l = 0, r = 1e9;
while(l < r)
{
LL mid = (l + r + 1) / 2;
if(mid * mid <= x)l = mid;
else r = mid - 1;
}
return r;
}
void solve()
{
LL x;
cin >> x;
if(x == 0)
{
cout << 0 << endl;
return;
}
int len = getlen(x);
for(int k = 0; k <= 19 - len ; k++)
{
LL p = 1;
for(int i = 1; i <= k ; i++)p = p * 10;
LL l = sl(x * p);
LL f = l * l;
for(int j = 1; j <= k ; j++)f /= 10;
if(f == x)
{
cout << l << endl;
return;
}
}
cout << "-1" << endl;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0), cout.tie(0);
int t;
cin >> t;
while (t--)
{
solve();
}
return 0;
}