bzoj 2301 HAOI2011 Problem b

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题目链接

BZOJ 2301 HAOI2011 Problem b

题解

$$\sum_{i=1}^n\sum_{j=1}^m[gcd(i,j)==k]$$
$$=\sum_{i=1}^{⌊ \dfrac{n}{k}⌋}\sum_{j=1}^{⌊\dfrac{m}{k}⌋}[gcd(i,j)==1]$$
利用函数f(x)表示x|(gcd(i,j))中ij的对数,那么原函数:
$$=\sum_{i=1}^{⌊ \dfrac{n}{k}⌋}\sum_{j=1}^{⌊\dfrac{m}{k}⌋}\sum_{d|gcd(i,j)}\mu(d)$$
$$=\sum_{d=1}^{min(\dfrac{n}{k},\dfrac{m}{k})} \mu(d)*\sum_{d|i,i\leq\dfrac{n}{k}}\sum_{d|j,j\leq\dfrac{m}{k}}1$$
$$\sum_{d=1}^{min(⌊\dfrac{n}{k}⌋,⌊\dfrac{m}{k}⌋)} \mu(d)*⌊\dfrac{n}{k}⌋*⌊\dfrac{m}{k}⌋)$$
$⌊\dfrac{n}{k}⌋$最多只有$2\sqrt{n}$个取值,预处理$\mu(d)$ $O(\sqrt{n})$回答

代码

#include<cstdio>
#include<cstring>
#include<algorithm>
typedef long long LL;
const int maxn=50007;
inline int read() {
    int x=0;
    char c=getchar();
    while(c<'0'||c>'9')c=getchar();
    while(c>='0'&&c<='9')x=x*10+c-'0',c=getchar();
    return x;
}
int n,prime[maxn];
bool vis[maxn];long long mu[maxn];
void get_wrrj(){
    mu[1]=1;
    for(int i=2;i<=maxn-1;i++) {
        if(!vis[i]) prime[++prime[0]]=i,mu[i]=-1;
        for(int j=1;j<=prime[0]&&i*prime[j]<=maxn-1;j++){
            vis[i*prime[j]]=1;
            if(i%prime[j]==0){
                mu[i*prime[j]]=0;break;
            }
            mu[i*prime[j]]=-mu[i];
        }
    }
    for(int i=1;i<=maxn-1;i++) mu[i]+=mu[i-1];
}
LL calc(int n,int m,int k){
    n/=k;m/=k;
    if(n>m) std::swap(n,m);
    LL ans=0;int next=0;
    for(int i=1;i<=n;i=next+1) {
        next=std::min(n/(n/i),m/(m/i));
        ans+=(mu[next]-mu[i-1])*(n/i)*(m/i);
    }
    return ans;
}
int main() {
    get_wrrj();
    int T=read();
    for(int a,b,c,d,k;T--;){
        a=read();b=read();c=read();d=read();k=read();
        printf("%lld\n",calc(b,d,k)-calc(a-1,d,k)-calc(b,c-1,k)+calc(a-1,c-1,k));
    }
    return 0;
}