bzoj 2301 HAOI2011 Problem b

题目链接

BZOJ 2301 HAOI2011 Problem b

题解

[sum_{i=1}^nsum_{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,ileqdfrac{n}{k}}sum_{d|j,jleqdfrac{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(2sqrt{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_asd() {
    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_asd();
    int T=read();
    for(int a,b,c,d,k;T--;) {
        a=read();b=read();c=read();d=read();k=read();
        printf("%lld
",calc(b,d,k)-calc(a-1,d,k)-calc(b,c-1,k)+calc(a-1,c-1,k));
    }
    return 0;
}