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 + 10;
ll mu[maxn], check[maxn], prime[maxn];
void Euler() 
{
    int tot = 0;
    mu[1] = 1;
    check[1] = true;
    for (int i = 2; i < maxn; i++) {
        if (!check[i]) {
            prime[++tot] = i;
            mu[i] = -1;
        }
        for (int j = 1; j <= tot; j++) 
        {
            if ((ll)prime[j] * i >= maxn) break;
            check[i*prime[j]] = true;
            if (i%prime[j] == 0) 
            {
                mu[i*prime[j]] = 0;
                break;
            }
            else mu[i*prime[j]] = mu[i] * -1;
        }
    }
}
int a, b, c, d;
ll why(int l,int r)
{
    ll ans=0;
    for (int i=1;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-1)-why(a-1,d)+why(a-1,c-1);
}