題目鏈接

BZOJ 2301 HAOI2011 Problem b

題解

\[\sum_{i=1}^n\sum_{j=1}^m[gcd(i,j)==k]\]
\[=\sum_{i=1}^{? \dfrac{n}{k}?}\sum_{j=1}^{?\dfrac{m}{k}?}[gcd(i,j)==1]\]
利用函數f(x)表示x|(gcd(i,j))中ij的對數,那麽原函數:
\[=\sum_{i=1}^{? \dfrac{n}{k}?}\sum_{j=1}^{?\dfrac{m}{k}?}\sum_{d|gcd(i,j)}\mu(d)\]
\[=\sum_{d=1}^{min(\dfrac{n}{k},\dfrac{m}{k})} \mu(d)*\sum_{d|i,i\leq\dfrac{n}{k}}\sum_{d|j,j\leq\dfrac{m}{k}}1\]

代碼

#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;
}

bzoj 2301 HAOI2011 Problem b