bzoj 2820: YY的GCD

 

http://www.lydsy.com/JudgeOnline/problem.php?id=2820

 

推一波式子

$$\sum_{isprime(p)}\sum_{a=1}^n\sum_{b=1}^mgcd(a,b)==p$$

$$\sum_{isprime(p)}\sum_{a=1}^{\left \lfloor \frac{n}{p} \right\rfloor}\sum_{b=1}^{\left \lfloor \frac{m}{p} \right \rfloor}gcd(a,b)==1$$

$$\sum_{isprime(p)}\sum_{a=1}^{\left \lfloor \frac{n}{p} \right \rfloor}\sum_{b=1}^{\left \lfloor \frac{m}{p} \right \rfloor}\sum_{d|gcd(a,b)}\mu(d)$$

$$\sum_{isprime(p)}\sum_{a=1}^{\left \lfloor \frac{n}{p} \right \rfloor}\sum_{b=1}^{\left \lfloor \frac{m}{p} \right \rfloor}\sum_{d|a \land d|b}\mu(d)$$

$$\sum_{isprime(p)}\sum_{d=1}^{\left \lfloor \frac{n}{p}\right \rfloor}\mu(d){\left\lfloor\frac{n}{pd}\right\rfloor}{\left\lfloor \frac{m}{pd} \right \rfloor}$$

将pd设为k

$$\sum_{k=1}^{n}\sum_{isprime(p) \land p|k}\mu(\frac{k}{p}){\left \lfloor \frac{n}{k} \right \rfloor}{\left \lfloor \frac{m}{k} \right \rfloor}$$
$$\sum_{k=1}^{n}F(k){\left \lfloor\frac{n}{k}\right\rfloor}{\left \lfloor \frac{m}{k} \right \rfloor}$$

就可以线性筛处理F了

#include<cstdio>
#include<algorithm>

#define maxn 10000007
#define LL long long
LL x,y;
int n,prime[maxn],miu[maxn],cnt,f[maxn],sum[maxn];
bool v[maxn];
void get_pre() {
    miu[1]=1;
    for(int i=2;i<=maxn-7;i++) {
        if(!v[i]) {
            v[i]=true;prime[++cnt]=i;
            miu[i]=-1;f[i]=1;
        }
        for(int j=1;j<=cnt;j++) {
            if(i*prime[j]>maxn-7) break;
            v[i*prime[j]]=true;
            if(i%prime[j]==0) {
                miu[i*prime[j]]=0;
                f[i*prime[j]]=miu[i];
                break;
            }
            miu[i*prime[j]]=-miu[i];
            f[i*prime[j]]=miu[i]-f[i];
        }
    }
    for(int i=1;i<=maxn-7;i++) sum[i]=sum[i-1]+f[i];
}
int main() {
    get_pre();
    scanf("%d",&n);
    while(n--) {
        scanf("%lld%lld",&x,&y);       
        LL k=std::min(x,y),j,ans=0;
            for(LL i=1;i<=k;i=j+1) {
            j=std::min(x/(x/i),y/(y/i));
            ans+=(x/i)*(y/i)*(sum[j]-sum[i-1]);
        }
        printf("%lld\n",ans);
    }
    return 0;
}