(Description)
(nleq 10^{10})，求
[sum_{i=1}^nsum_{j=1}^ngcd(i,j) mod (1e9+7)
]
(Solution)
首先
[sum_{i=1}^nsum_{j=1}^ngcd(i,j)=sum_{d=1}^ndsum_{i=1}^nsum_{j=1}^n[gcd(i,j)=d]
]
注意不是(sum_{i=1}^nsum_{j=1}^ngcd(i,j)=sum_{d=1}^nsum_{i=1}^nsum_{j=1}^n[gcd(i,j)=d])！
[ egin{aligned}
sum_{i=1}^nsum_{j=1}^ngcd(i,j)&=sum_{d=1}^ndsum_{i=1}^nsum_{j=1}^n[gcd(i,j)=d]\
&=sum_{d=1}^ndsum_{i=1}^{lfloorfrac{n}{d}
floor}sum_{j=1}^{lfloorfrac{n}{d}
floor}[gcd(i,j)=1]\
&=sum_{d=1}^ndsum_{i=1}^{lfloorfrac{n}{d}
floor}mu(i)lfloorfrac{lfloorfrac{n}{d}
floor}{i}
floorlfloorfrac{lfloorfrac{n}{d}
floor}{i}
floor\
&=sum_{d=1}^ndsum_{d|t,tleq n}mu(frac{t}{d})(lfloorfrac{n}{t}
floor)^2\
&=sum_{t=1}^n(lfloorfrac{n}{t}
floor)^2sum_{d|t}dmu(frac{t}{d})\
&=sum_{t=1}^n(lfloorfrac{n}{t}
floor)^2varphi(t)
end{aligned}
]
然后就可以杜教筛了。
最后一步用到$$sum_{d|n}frac{n}{d}mu(d)=varphi(n)$$
证明：
首先有 (sum_{d|n}varphi(d)=n)（不证了）.
设 (f(n)=n)，则(f(n)=sum_{d|n}varphi(d)).
那么 (varphi(n)=sum_{d|n}f(frac{n}{d})mu(d))
即 (sum_{d|n}frac{n}{d}mu(d)=varphi(n)).
//前缀和不要忘取模。。
#include<cstdio>
#include<cstring>
typedef long long LL;
const int N=4500000,mod=1e9+7;
int P[N>>2],cnt,phi[N+3];
LL n,sum[N+3],sum2[100000];
bool Not_P[N+3];
void Init()
{
phi[1]=1;
for(int i=2;i<=N;++i)
{
if(!Not_P[i]) P[++cnt]=i,phi[i]=i-1;
for(int j=1;j<=cnt&&i*P[j]<=N;++j)
{
Not_P[i*P[j]]=1;
if(i%P[j]) phi[i*P[j]]=phi[i]*(P[j]-1);
else {phi[i*P[j]]=phi[i]*P[j]; break;}
}
}
for(int i=1;i<=N;++i) sum[i]=sum[i-1]+phi[i], sum[i]>=mod?sum[i]-=mod:0;
}
LL FP(LL x,LL k)
{
LL t=1;
for(;k;k>>=1,x=x*x%mod)
if(k&1) t=t*x%mod;
return t;
}
const LL inv2=FP(2,mod-2);
LL Calc(LL x)
{
if(x<=N) return sum[x];
else if(~sum2[n/x]) return sum2[n/x];
LL t=x%mod, res=t*(t+1)%mod*inv2%mod;
for(LL i=2,las;i<=x;i=las+1)
las=x/(x/i),(res-=(las-i+1)*Calc(x/i)%mod)%=mod;
return sum2[n/x]=(res+mod)%mod;
}
int main()
{
Init();//scanf("%I64d",&n);
scanf("%lld",&n);
memset(sum2,0xff,sizeof sum2);
LL res=0;
for(LL i=1,las,t;i<=n;i=las+1)
las=n/(n/i), t=n/i%mod, (res+=t*t%mod*(Calc(las)-Calc(i-1)+mod)%mod)%=mod;
printf("%lld",res);
return 0;
}//9800581876