bzoj_2693: jzptab

Posted 2020-10-11 A_LEAF

跟那个Crash一样

不过是进行了优化

后面可以线筛

不互质的时候，i*prime[j]的因数mu变成了0，所以只需要f[i*prime[j]]=f[i]*prime[j]

 

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
inline int read()
{
    char q=getchar();int ans=0;
    while(q<‘0‘||q>‘9‘)q=getchar();
    while(q>=‘0‘&&q<=‘9‘){ans=ans*10+q-‘0‘;q=getchar();}
    return ans;
}
const int mod=100000009;//多打了一个0 
const int N=10000006;

int prime[N],cnt;
bool he[N];
int mu[N];
ll ji[N];
int T,n,m;

void chu()
{
    mu[1]=1;ji[1]=1;
    for(int i=2;i<N;++i)
    {
        if(!he[i])
        {
            prime[++cnt]=i;
            mu[i]=-1;
            ji[i]=(((ll)i-(ll)i*i)%mod+mod)%mod;
        }
        for(int j=1;j<=cnt&&prime[j]*i<N;++j)
        {
            he[i*prime[j]]=1;
            if(i%prime[j]==0)
            {
                mu[i*prime[j]]=0;
                ji[i*prime[j]]=ji[i]*prime[j]%mod;
                // 有>2个prime[j]因子的mu是0,所以多出来的只有f[i]*prime[j] 
                break;
            }
            mu[i*prime[j]]=-mu[i];
            ji[i*prime[j]]=ji[i]*ji[prime[j]]%mod;
        }
    }
    
    for(int i=1;i<N;++i)
      ji[i]=(ji[i]+ji[i-1])%mod;
}

ll sum(ll x,ll y)
{
    ll t1,t2;
    t1=(x+1)*x/2%mod;
    t2=(y+1)*y/2%mod;
    return t1*t2%mod;
}

ll work()
{
    if(n>m)
        swap(n,m);
    ll ans=0;
    int nx;
    for(int i=1;i<=n;i=nx+1)
    {
        nx=min( n/(n/i),m/(m/i) );
        ans=(ans+sum(n/i,m/i)*(ji[nx]-ji[i-1]+mod)%mod)%mod;
    }
    return (ans+mod)%mod;
}

int main(){
    
    //freopen("in.in","r",stdin);
    //freopen("bzoj_2693.in","r",stdin);
    //freopen("bzoj_2693.out","w",stdout);
    
    chu();
    T=read();
    while(T--)
    {
        n=read();m=read();
        //printf("n=%d m=%d\n",n,m);
        printf("%lld\n",work());
    }
}
    

AA