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)

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 const ll maxn = 1e7 + ;
 ll mu[maxn], check[maxn], prime[maxn];
 void Euler()
 {
     int tot = ;
     mu[] = ;
     check[] = true;
     for (int i = ; i < maxn; i++) {
         if (!check[i]) {
             prime[++tot] = i;
             mu[i] = -;
         }
         for (int j = ; j <= tot; j++)
         {
             if ((ll)prime[j] * i >= maxn) break;
             check[i*prime[j]] = true;
             if (i%prime[j] == )
             {
                 mu[i*prime[j]] = ;
                 break;
             }
             else mu[i*prime[j]] = mu[i] * -;
         }
     }
 }
 int a, b, c, d;
 ll why(int l,int r)
 {
     ll ans=;
     for (int i=;i<=min(l, r);i++)
         ans+=(ll)(l/i)*(r/i)*mu[i];
     return ans;
 }
 int main()
 {
     Euler();
     cin >> a >> b >> c >> d;
     cout<<why(b,d)-why(b,c-)-why(a-,d)+why(a-,c-);
 }