bzoj2693: jzptab

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@jzq233jzq
没看过2154的右转

答案=∑d=1∑i=1ndi∑j=1mdj(gcd(i,j)==1)

然后再化开。

=∑d=1d∑b=1μ(b)∗b2∗∑i=1ndbi∑j=1mdbj

设D=db
则

=∑D=1∑i=1nDi∑j=1mDj∗∑D|bDbμ(b)∗b2

然后设f(x)=j∗∑D|bDbmu(b)∗b2
可以发现f(x)是积性函数。
线性筛一发+分块枚举D水过。

#include<cmath>#include<cstdio>#include<cstring>#include<cstdlib>#include<iostream>#include<algorithm>#define ll long longusing namespace std;const ll mo=100000009;const int N=10000005;ll h[N],ans;int pri[N/10],T,n,m,tot;bool fl[N];ll sum(ll x,ll y){    ll s1=x*(x+1)/2%mo;    ll s2=y*(y+1)/2%mo;    return s1*s2%mo; }int main(){    h[1]=1;    for (int i=2;i<N;i++){        if (!fl[i]){            pri[++tot]=i;            h[i]=(i-1ll*i*i%mo+mo)%mo;        }        for (int j=1;pri[j]*i<N&&j<=tot;j++){            fl[pri[j]*i]=1;            if (i%pri[j]==0){                h[pri[j]*i]=pri[j]*h[i]%mo;                break;            }            h[pri[j]*i]=h[pri[j]]*h[i]%mo;        }    }    for (int i=1;i<N;i++)        h[i]=(h[i]+h[i-1])%mo;    scanf("%d",&T);    while (T--){        scanf("%d%d",&n,&m);        ans=0;        if (n>m) swap(n,m);        for (int i=1,j;i<=n;i=j+1){            j=min(n/(n/i),m/(m/i));            ans=(ans+sum(n/i,m/i)*(h[j]-h[i-1])%mo+mo)%mo;         }        printf("%lld\n",ans);    }}