• 洛谷 P2257 YY的GCD


    Code:

    #pragma GCC optimize(3)
    #pragma GCC optimize(2)
    #include <map>
    #include <set>
    // #include <array>
    #include <cmath>
    #include <queue>
    #include <stack>
    #include <vector>
    #include <cstdio>
    #include <cstring>
    #include <sstream>
    #include <iostream>
    #include <stdlib.h>
    #include <algorithm>
    // #include <unordered_map>
    
    using namespace std;
    
    typedef long long ll;
    typedef pair<int, int> PII;
    
    #define Time (double)clock() / CLOCKS_PER_SEC
    
    #define sd(a) scanf("%d", &a)
    #define sdd(a, b) scanf("%d%d", &a, &b)
    #define slld(a) scanf("%lld", &a)
    #define slldd(a, b) scanf("%lld%lld", &a, &b)
    
    const int N = 1e7 + 20;
    const int M = 1e5 + 20;
    const int mod = 1e9 + 7;
    const double eps = 1e-6;
    
    ll cnt = 0, primes[N], phi[N], sum[N] = {0}, mu[N];
    map<ll, ll> su;
    bool st[N];
    
    int n, m, p, k;
    
    void get(ll n){
        // phi[1] = 1;
        mu[1] = 1;
        for(ll i = 2; i <= n; i ++){
            if(!st[i]){
                primes[cnt ++] = i;
                // phi[i] = i - 1;
                mu[i] = -1;
            }
            for(int j = 0; primes[j] <= n / i; j ++){
                st[i * primes[j]] = true;
                if(i % primes[j] == 0){
                    // phi[i * primes[j]] = primes[j] * phi[i];
                    mu[i * primes[j]] = 0;
                    break;
                }
                // phi[i * primes[j]] = (primes[j] - 1) * phi[i];
                mu[i * primes[j]] = - mu[i];
            }
        }
    }
    
    void init(int n){
        for(int i = 0; i < cnt; i ++){
            int p = primes[i];
            for(int j = p; j <= n; j += p){
                sum[j] += mu[j / p];
            }
        }
        for(int i = 2; i <= n; i ++){
            sum[i] += sum[i - 1];
        }
    }
    
    
    void solve()
    {
        sdd(n, m);
        ll ans = 0;
        if(n > m) swap(n, m);
        for(int l = 1, r; l <= n; l = r + 1){
            r = min(n / (n / l), m / (m / l));
            ll res = (ll)(n / l) * (m / l), mid = sum[r] - sum[l - 1];
            ans += res * mid;
        }
    
        printf("%lld
    ", ans);
    
    }
    
    int main()
    {
    #ifdef ONLINE_JUDGE
    #else
        freopen("/home/jungu/code/in.txt", "r", stdin);
        // freopen("/home/jungu/code/out.txt", "w", stdout);
        // freopen("/home/jungu/code/practice/out.txt","w",stdout);
    #endif
        // ios::sync_with_stdio(false);
        cin.tie(0), cout.tie(0);
    
        int T = 1;
        sd(T);
        get(10000000);
        // k = (int)pow(10000000, 0.666667);
        init(10000000);
        // int cas = 1;
        while (T--)
        {
            // printf("Case #%d:", cas++);
            solve();
        }
    
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/jungu/p/13526257.html
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