Bzoj2818: Gcd

最新推荐文章于 2019-05-08 21:43:56 发布

原创最新推荐文章于 2019-05-08 21:43:56 发布 · 256 阅读

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本文介绍了一道涉及数论的题目及其解决方案，通过定义函数f(i)和g(i)来简化问题，并利用质数的特性进行求解。文章还展示了如何使用前缀和与数论分块的方法提高计算效率。

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题面

戳我

Sol

~~傻逼题~~
$原式=\sum_{p质数}^{n}\sum_{i=1}^{\lfloor\frac{n}{p}\rfloor}\sum_{j=1}^{\lfloor\frac{m}{p}\rfloor}[gcd(i, j)==1]$
$设f(i) = \sum_{i=1}^{\lfloor\frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor\frac{m}{d}\rfloor}[gcd(i, j)==1]$
$g(i) = \sum_{i|d} f(d) = \lfloor\frac{\lfloor\frac{n}{d}\rfloor}{i}\rfloor\lfloor\frac{\lfloor\frac{m}{d}\rfloor}{j}\rfloor$
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\n", ans);
    return 0;
}