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  • Bzoj2818: Gcd

    题面

    戳我

    Sol

    傻逼题
    (原式=sum_{p质数}^{n}sum_{i=1}^{lfloorfrac{n}{p} floor}sum_{j=1}^{lfloorfrac{m}{p} floor}[gcd(i, j)==1])
    (设f(i) = sum_{i=1}^{lfloorfrac{n}{d} floor}sum_{j=1}^{lfloorfrac{m}{d} floor}[gcd(i, j)==1])
    (g(i) = sum_{i|d} f(d) = lfloorfrac{lfloorfrac{n}{d} floor}{i} floorlfloorfrac{lfloorfrac{m}{d} floor}{j} floor)
    质数就把质数的bool求前缀和就好了,数论分块

    # include <bits/stdc++.h>
    # define RG register
    # define IL inline
    # define Fill(a, b) memset(a, b, sizeof(a))
    using namespace std;
    typedef long long ll;
    const int _(1e7 + 1);
    
    IL ll Read(){
        char c = '%'; ll x = 0, z = 1;
        for(; c > '9' || c < '0'; c = getchar()) if(c == '-') z = -1;
        for(; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
        return x * z;
    }
    
    int prime[_], mu[_], num, s[_];
    bool isprime[_];
    
    IL void Prepare(){
    	isprime[1] = 1; mu[1] = 1;
    	for(RG int i = 2; i < _; ++i){
    		if(!isprime[i]) prime[++num] = i, mu[i] = -1;
    		for(RG int j = 1; j <= num && i * prime[j] < _; ++j){
    			isprime[i * prime[j]] = 1;
    			if(i % prime[j])  mu[i * prime[j]] = -mu[i];
    			else{  mu[i * prime[j]] = 0; break;	 }
    		}
    		mu[i] += mu[i - 1]; s[i] = s[i - 1] + (!isprime[i]);
    	}
    }
    
    IL ll Calc(RG ll n){
    	RG ll f = 0;
    	for(RG ll i = 1, j; i <= n; i = j + 1) j = n / (n / i), f += 1LL * (mu[j] - mu[i - 1]) * (n / i) * (n / i);
    	return f;
    }
    
    int main(RG int argc, RG char *argv[]){
    	Prepare();
    	RG int n = Read(); RG ll ans = 0;
    	for(RG ll d = 1, j; d <= n; d = j + 1) j = n / (n / d), ans += 1LL * (s[j] - s[d - 1]) * Calc(n / d);
    	printf("%lld
    ", ans);
    	return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/cjoieryl/p/8260773.html
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