Gym - 101982B 2018-2019 ACM-ICPC Pacific Northwest Regional Contest (Div. 1) B. Coprime Integers Mobius+容斥 ab间gcd(x,y)=1的对数

题面

题意:给你 abcd(1e7),求a<=x<=b,c<=y<=d的,gcd(x,y)=1的数量

题解:经典题目,求从1的到n中选x,从1到m中选y的,gcd(x,y)=k的可以看hdu1695

        ask(x,y)表示从1到n中选n,从1到m中选y的话

        ans=ask(b,d)-ask(b,c-1)-ask(a-1,d)+ask(a-1,c-1)

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const ll maxn = 1e7 + 10;
 5 ll mu[maxn], check[maxn], prime[maxn];
 6 void Euler() 
 7 {
 8     int tot = 0;
 9     mu[1] = 1;
10     check[1] = true;
11     for (int i = 2; i < maxn; i++) {
12         if (!check[i]) {
13             prime[++tot] = i;
14             mu[i] = -1;
15         }
16         for (int j = 1; j <= tot; j++) 
17         {
18             if ((ll)prime[j] * i >= maxn) break;
19             check[i*prime[j]] = true;
20             if (i%prime[j] == 0) 
21             {
22                 mu[i*prime[j]] = 0;
23                 break;
24             }
25             else mu[i*prime[j]] = mu[i] * -1;
26         }
27     }
28 }
29 int a, b, c, d;
30 ll why(int l,int r)
31 {
32     ll ans=0;
33     for (int i=1;i<=min(l, r);i++)
34         ans+=(ll)(l/i)*(r/i)*mu[i];
35     return ans;
36 }
37 int main() 
38 {
39     Euler();
40     cin >> a >> b >> c >> d;
41     cout<<why(b,d)-why(b,c-1)-why(a-1,d)+why(a-1,c-1);
42 }