P2257 YY的GCD

链接

思路

 

题目要求
$\sum\limits_{i=1}^n\sum\limits_{j=1}^m [(i,j)为素数]$

枚举一个素数，考虑它的贡献

$\sum\limits_p\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{m} [(i,j)=p]$

后面的利用bzoj 1101的做法。

$\sum\limits_p\sum\limits_{i=1}^{\lfloor\frac{n}{p}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{m}{p}\rfloor} [(i,j)=1]$

$\sum\limits_p \sum\limits_{d=1}^{min(\lfloor\frac{n}{p}\rfloor,\lfloor\frac{m}{p}\rfloor)}μ(d)\lfloor\frac{n}{pd}\rfloor\lfloor\frac{m}{pd}\rfloor$

设T=pd，枚举T，再枚举T的素因数p，那么d也就等于T/p

$\sum\limits_T\sum\limits_{p|T}μ(\frac{T}{p})\lfloor\frac{n}{T}\rfloor\lfloor\frac{m}{T}\rfloor$

后面那一块只跟前面的T有关，所以可以提前。

$\sum\limits_T\lfloor\frac{n}{T}\rfloor\lfloor\frac{m}{T}\rfloor（\sum\limits_{p|T}μ(\frac{T}{p})）$

将后面的预处理出来。
参考popoqqq的课件，之间枚举一个素数，然后更新p*1,p*2...。

复杂度：1/1+1/2+1/3+...+1/n=O(logn)

 

代码

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>

using namespace std;

const int N = 10000000;
typedef long long LL;

bool noprime[N+5];
int pri[N+5],tot,mu[N+5],g[N+5];

inline int read() {
    int x = 0,f = 1;char ch = getchar();
    for (; !isdigit(ch); ch=getchar()) if(ch=='-') f=-1;
    for (; isdigit(ch); ch=getchar()) x=x*10+ch-'0';
    return x * f;
}
void init() {
    mu[1] = 1; //-
    for (int i=2; i<=N; ++i) {
        if (!noprime[i]) pri[++tot] = i,mu[i] = -1;
        for (int j=1; j<=tot&&i*pri[j]<=N; ++j) {
            noprime[i*pri[j]] = true;
            if (i % pri[j] == 0) {mu[i*pri[j]] = 0;break;}
            mu[i*pri[j]] = -mu[i];
        }
    }
    for (int j=1; j<=tot; ++j) 
        for (int i=pri[j]; i<=N; i+=pri[j])
            g[i] += mu[i/pri[j]];
    for (int i=1; i<=N; ++i) g[i] += g[i-1];
}
void work(int n,int m) {
    LL ans = 0;
    int c = min(n,m),p;
    for (int i=1; i<=c; i=p+1) {
        p = min(n/(n/i),m/(m/i));
        ans += 1ll*(g[p]-g[i-1])*(n/i)*(m/i);
    }
    printf("%lld\n",ans);
}
int main() {
    init();
    int Case = read();
    while (Case--) {
        int n = read(),m = read();
        work(n,m);
    }
    return 0;
}