莫比乌斯反演模板--Gym 101982B

Posted 2021-12-03 shuizhidao

莫比乌斯反演模板--Gym 101982B

题意

给你a,b,c,d四个数，求[a,b]与[c,d]，互质的数的对数

代码

#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <cmath>
#include <sstream>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <iomanip>
#include <stack>

using namespace std;

typedef long long LL;
const int INF = 0x3f3f3f3f;
const int N = 1e7 + 100;
const int MOD = 1e9 + 9;

#define lson l, m, rt << 14
#define rson m + 1, r, rt << 1 | 1
#define F(i, l, r) for(int i = l;i <= (r);++i)
#define RF(i, l, r) for(int i = l;i >= (r);--i)

int prime[N], vis[N], cnt, mu[N], sum[N];
void Isprime(int n)
{
    mu[1] = 1;
    for(int i = 2;i <= n;++i)
    {
        if(!vis[i]) {prime[cnt++] = i;mu[i] = -1;}
        for(int j = 0;j < cnt && i * prime[j] <= n;++j)
        {
            vis[i * prime[j]] = 1;
            if(i % prime[j] == 0)
                break;
            mu[i * prime[j]] = -mu[i];
        }
    }
    for(int i = 1;i <= n;++i)
        sum[i] = sum[i - 1] + mu[i];
}

LL a, b, c, d;
LL solve(LL n, LL m)
{
    if(n > m) swap(n, m);
    LL ans = 0;
    for(int i = 1, last;i <= n;i = last + 1)
    {
        last = min(n / (n / i), m / (m / i));
        ans += (sum[last] - sum[i - 1]) * (n / i) * (m / i);
    }
    return ans;
}

int main()
{
    Isprime(1e7+5);
    cin >> a >> b >> c >> d;
    cout << solve(b, d) - solve(a - 1, d) - solve(b, c - 1) + solve(a - 1, c - 1);
    return 0;
}