题目链接：
https://www.lydsy.com/JudgeOnline/problem.php?id=2440
题目大意：
求第k个无平方因子的数
思路：
二分答案x，求1-x中有多少个平方因子的数
可以在根号x的范围内求出来
1 #include<bits/stdc++.h> 2 #define IOS ios::sync_with_stdio(false);//不可再使用scanf printf 3 #define Max(a, b) ((a) > (b) ? (a) : (b))//禁用于函数，会超时 4 #define Min(a, b) ((a) < (b) ? (a) : (b)) 5 #define Mem(a) memset(a, 0, sizeof(a)) 6 #define Dis(x, y, x1, y1) ((x - x1) * (x - x1) + (y - y1) * (y - y1)) 7 #define MID(l, r) ((l) + ((r) - (l)) / 2) 8 #define lson ((o)<<1) 9 #define rson ((o)<<1|1) 10 #pragma comment(linker, "/STACK:102400000,102400000")//栈外挂 11 using namespace std; 12 inline int read() 13 { 14 int x=0,f=1;char ch=getchar(); 15 while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();} 16 while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} 17 return x*f; 18 } 19 20 typedef long long ll; 21 const int maxn = 100000 + 10; 22 const int maxm = 100000 + 10; 23 const int MOD = 1000000007;//const引用更快，宏定义也更快 24 const ll INF = 1644934082; 25 const double eps = 1e-6; 26 bool not_prime[maxn]; 27 int prime[maxn]; 28 int Mob[maxn]; 29 void Mobius_sieve(int n) 30 { 31 int tot = 0; 32 not_prime[1] = 1; 33 Mob[1] = 1; 34 for(int i = 2; i <= n; i++) 35 { 36 if(!not_prime[i])prime[tot++] = i, Mob[i] = - 1; 37 for(int j = 0; j < tot && 1LL * prime[j] * i <= n; j++) 38 { 39 not_prime[prime[j] * i] = 1;//每个合数x由它最小素因子prime[j]筛掉 40 Mob[i * prime[j]] = (i % prime[j] ? -Mob[i]: 0); 41 if(i % prime[j] == 0)break;//如果i % prime[j] == 0,不停止循环 42 //那么接下来将用prime[j+1]筛去i*prime[j+1]，但实际上应该用prime[i]筛去,因为i%prime[j]==0 43 } 44 } 45 } 46 ll judge(ll m) 47 { 48 ll sum = 0; 49 for(ll i = 1; i * i <= m; i++) 50 { 51 ll tmp = m / i / i; 52 sum += Mob[i] * tmp; 53 } 54 return sum; 55 } 56 int main() 57 { 58 Mobius_sieve(100000); 59 int T; 60 scanf("%d", &T); 61 while(T--) 62 { 63 ll k; 64 scanf("%lld", &k); 65 ll l = 1, r = INF; 66 ll ans; 67 while(l <= r) 68 { 69 ll m = (l + r) / 2; 70 if(judge(m) >= k)ans = m, r = m - 1; 71 else l = m + 1; 72 } 73 printf("%lld ", ans); 74 } 75 return 0; 76 }