HDU 5778 ABS (Prime, violence)

Test instructions: Given a number x, the positive integer y≥2y\geq 2y≥2, so that the following conditions are met: 1.y-x the absolute value of the minimum 2.y of the mass-factor decomposition of each factorization is exactly 2 times.

Analysis: Because each factorization in the decomposition of Y-factor is 2 times, then Y is a complete square number, set Y=z*z, the title can be converted into Z, so that each factorization appears 1 times. We can violently enumerate Z, check if z meets the requirements,

Obviously, when Z is a prime number that is compliant, the enumerator of Z is available by the prime numbers theorem, the Logn level complexity O (n4logn2\sqrt[4]{n}log\sqrt[2]{n}.

Pay attention to y>=2.

The code is as follows:

#include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream > #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector > #include <map> #include <cctype>using namespace std; typedef long Long Ll;typedef pair<int, int> p;c onst int inf = 0x3f3f3f3f;const double inf = 0x3f3f3f3f3f3f3f;const double EPS = 1e-8;const int maxn = 1e5 + 5;const int D R[] = {0, 0,-1, 1};const int dc[] = {-1, 1, 0, 0};int N, m;inline bool is_in (int r, int c) {return R >= 0 &&amp ; R < n && C >= 0 && C < m;}    inline bool Solve (LL x) {int n = (int) sqrt (x+0.5);        for (int i = 2; I <= n; ++i) if (x% i = = 0) {x/= i;    if (x% i = = 0) return true; } return false;}    int main () {//Ios::sync_with_stdio (FALSE);  int T, q;    Cin >> T;        while (t--) {LL x;        CIN >> X;        ll T1 = (ll) sqrt (x*1.0); LL t2 = t1+1;        while (T1 > 1 && Solve (t1))--t1;        while (Solve (T2)) ++t2;        LL ans = ABS (t2*t2-x);        if (T1 >= 2) ans = min (ans, ABS (t1*t1-x));    cout << ans << endl; } return 0;}

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HDU 5778 ABS (Prime, violence)