Educational Codeforces Round 50 (Rated for Div. 2) F - Relatively Prime Powers（数学+容斥）

题目链接：http://codeforces.com/contest/1036/problem/F

题意：

题解：求在[2,n]中,x != a ^ b（b >= 2 即为gcd）的个数，那么实际上就是要求x = a ^ b的个数，然后用总数减掉就好了，答案即为。（pow会丢失精度，学习避免精度丢失的方法）

 #include <bits/stdc++.h>
 using namespace std;
 #define ll long long
 #define LL __int128
 #define ull unsigned long long
 #define mst(a,b) memset((a),(b),sizeof(a))
 #define mp(a,b) make_pair(a,b)
 #define pi acos(-1)
 #define pii pair<int,int>
 #define pb push_back
 const int INF = 0x3f3f3f3f;
 const double eps = 1e-;
 const int MAXN = 1e5 + ;
 const int MAXM = 2e6 + ;
 const ll mod = 1e9 + ;

 bool check[];
 int prime[];
 int mu[];

 void Moblus(int N) {
     mst(check, false);
     mu[] = ;
     int tot = ;
     for(int i = ; i <= N; i++) {
         if(!check[i]) {
             prime[tot++] = i;
             mu[i] = -;
         }
         for(int j = ; j < tot; j++) {
             if(i * prime[j] > N) {
                 break;
             }
             check[i * prime[j]] = true;
             if(i % prime[j] == ) {
                 mu[i * prime[j]] = ;
                 break;
             } else {
                 mu[i * prime[j]] = -mu[i];
             }
         }
     }
 }

 int main()
 {
 #ifdef local
     freopen("data.txt", "r", stdin);
 //    freopen("data.txt", "w", stdout);
 #endif
     Moblus();
     int t;
     scanf("%d",&t);
     while(t--) {
         ll n;
         scanf("%lld",&n);
         ll ans = ;
         for(int i = ; i <= ; i++) {
             if(!mu[i]) continue;
             ll raiz = round(pow(n, 1.0 / i));
             while(pow((long double)raiz, i) > (long double)n) raiz--;
             while(pow((long double)raiz + , i) <= (long double)n) raiz++;
             raiz--;
             ans += (ll)mu[i] * raiz;
         }
         printf("%lld\n",n -  + ans);
     }
     return ;
 }