洛谷P2257 YY的GCD

Dear 丶 2022-01-07 05:21 178阅读 0赞

# 莫比乌斯反演 #

还是把gcd换成莫比乌斯函数，在最外层枚举素数p，对于每个素数p，gcd莫比乌斯反演

#include <bits/stdc++.h>
    #define INF 0x3f3f3f3f
    #define full(a, b) memset(a, b, sizeof a)
    #define FAST_IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
    using namespace std;
    typedef long long ll;
    inline int lowbit(int x){ return x & (-x); }
    inline int read(){
        int X = 0, w = 0; char ch = 0;
        while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
        while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
        return w ? -X : X;
    }
    inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }
    inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
    template<typename T>
    inline T max(T x, T y, T z){ return max(max(x, y), z); }
    template<typename T>
    inline T min(T x, T y, T z){ return min(min(x, y), z); }
    template<typename A, typename B, typename C>
    inline A fpow(A x, B p, C lyd){
        A ans = 1;
        for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
        return ans;
    }
    const int N = 10000005;
    int _, n, m, tot, mo[N], prime[N], f[N];
    bool vis[N];
    ll sum[N];
    void mobius(){
        mo[1] = 1;
        for(int i = 2; i < N; i ++){
            if(!vis[i]) prime[++tot] = i, mo[i] = -1;
            for(int j = 1; j <= tot && i * prime[j] < N; j ++){
                vis[i * prime[j]] = true;
                if(i % prime[j] == 0){
                    mo[i * prime[j]] = 0;
                    break;
                }
                else mo[i * prime[j]] = -mo[i];
            }
        }
        for(int i = 1; i <= tot; i ++){
            for(int j = 1; j * prime[i] < N; j ++) f[j * prime[i]] += mo[j];
        }
        for(int i = 2; i < N; i ++) sum[i] = sum[i - 1] + 1LL * f[i];
    }
    
    int main(){
    
        mobius();
        for(_ = read(); _; _ --){
            n = read(), m = read();
            ll ans = 0;
            for(int l = 1, r; l <= min(n, m); l = r + 1){
                r = min(n / (n / l), m / (m / l));
                ans += 1LL * (n / l) * (m / l) * (sum[r] - sum[l - 1]);
            }
            printf("%lld\n", ans);
        }
        return 0;
    }

转载于:https://www.cnblogs.com/onionQAQ/p/11013047.html