HDU 3826 squarefree Number (prime) __ Number theory

Description

In mathematics, a squarefree number is one which are divisible by no perfect squares, except 1. For example, Square-free but, as it is divisible by 9 = 3^2. Now you are need to determine whether an integer is Squarefree or.

Input

The ' The ' contains an integer T indicating the number of test cases.

For each test case, the there is a single line contains an integer N.

Technical Specification 1 <= T <= 2 <= N <= 10^18

Output

For each test case, output the case number A. Then output "Yes" If N is Squarefree, "No" otherwise.

Sample Input

2
75

Sample Output

Case 1:yes Case
2:no

the

To determine whether a number is "no square number" (Is this translation?) ）

train of Thought

Obviously the data range of 1018 10^{18} does not allow us to try each.

Considering the decomposition of N to mass factor, we know that if a prime number is greater than 2, it means that it is not a square number.

After the screening of all the 10^6 below 106, if n>106 n>10^6, it means that it contains up to two 106 10^6 above the quality factor, and then squared to determine whether the two qualitative factors are the same.

AC Code

 #include <bits/stdc++.h> using namespace std; typedef __int64 LL; const int MAXN = 1E6+10;
BOOL VISIT[MAXN];
int PRIME[MAXN];
int tot;
LL N;
    void Getprime () {tot=0;
    memset (visit,0,sizeof (visit));
            for (int i=2; i<maxn; i++) {if (!visit[i)) {prime[tot++]=i;
        for (int j=i+i; j<maxn; j+=i) visit[j]=1; BOOL Solve () {for (int i=0; i<tot; i++) {if (n%prime[i]==0) {n/=prime[i
            ];
        if (n%prime[i]==0) return false;
        } if (n>=1e6) {LL tmp = sqrt (n) + 0.5;
    if (tmp*tmp==n) return false;
return true;
    int main () {getprime ();
    int T;
    cin>>t;
        for (int ti=1; ti<=t; ti++) {cin>>n;
        cout<< "Case" <<ti<< ":"; cout<< (Solve ()? "
    Yes ":" No ") <<endl;
return 0; }