以后这种题能用phi的就不要用mu…mu往往会带着个ln然后被卡常致死
把题目要求转换为前缀和相减的形式，写出来大概是要求这样一个式子：
[sum_{i=1}^{n}sum_{j=1}^{i}frac{j}{gcd(i,d)}
]
注意j的限制是i
[sum_{d=1}^{n}sum_{i=1}^{n}sum_{j=1}^{i}[gcd(i,j)==d]frac{j}{d}
]
[sum_{d=1}^{n}sum_{i=1}^{left lfloor frac{n}{d}
ight
floor}sum_{j=1}^{i}[gcd(i,j)==1]frac{jd}{d}
]
[sum_{d=1}^{n}sum_{i=1}^{left lfloor frac{n}{d}
ight
floor}sum_{j=1}^{i}[gcd(i,j)==1]j
]
然后有一个打表找规律发现的式子：
[sum_{i=1}^{n}sum_{j=1}^{i}[gcd(i,j)==1]i*j=sum_{i=1}^{n}i*frac{phi(i)*i+[i==1]}{2}
]
于是原式可转化为：
[sum_{d=1}^{n}sum_{i=1}^{left lfloor frac{n}{d}
ight
floor}frac{phi(i)*i+[i==1]}{2}
]
[frac{sum_{d=1}^{n}sum_{i=1}^{left lfloor frac{n}{d}
ight
floor}phi(i)*i}{2}+frac{n}{2}
]
先不考虑后面的加和下面的除二，于是要求的就是：
[sum_{d=1}^{n}sum_{i=1}^{left lfloor frac{n}{d}
ight
floor}phi(i)*i
]
( sum_{i=1}^{left lfloor frac{n}{d} ight floor}phi(i)*i )的部分显然可以用杜教筛处理，然后拒绝算时间复杂度。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const long long N=1000005,m=1000000,inv6=166666668,inv2=500000004,mod=1e9+7;
long long a,b,n,phi[N],q[N],tot,ha[N];
bool v[N];
long long clc1(long long x)
{
return x*(x+1)%mod*inv2%mod;
}
long long clc2(long long x)
{
return x*(x+1)%mod*(2*x+1)%mod*inv6%mod;
}
long long slv(long long x)
{
if(x<=m)
return phi[x];
if(ha[n/x])
return ha[n/x];
long long re=clc2(x);
for(long long i=2,la;i<=x;i=la+1)
{
la=x/(x/i);
re=(re-(clc1(la)-clc1(i-1))%mod*slv(x/i)%mod)%mod;
}
return ha[n/x]=re;
}
long long wk(long long x)
{
n=x;
memset(ha,0,sizeof(ha));
long long re=0ll;
for(long long i=1,la;i<=x;i=la+1)
{
la=x/(x/i);
re=(re+(la-i+1)*slv(x/i)%mod)%mod;
}
return (re+x)*inv2%mod;
}
int main()
{
phi[1]=1;
for(long long i=2;i<=m;i++)
{
if(!v[i])
{
q[++tot]=i;
phi[i]=i-1;
}
for(long long j=1;j<=tot&&i*q[j]<=m;j++)
{
long long k=i*q[j];
v[k]=1;
if(i%q[j]==0)
{
phi[k]=phi[i]*q[j];
break;
}
phi[k]=phi[i]*(q[j]-1);
}
}
for(long long i=1;i<=m;i++)
phi[i]=(phi[i-1]+phi[i]*i%mod)%mod;
scanf("%lld%lld",&a,&b);
printf("%lld
",((wk(b)-wk(a-1))%mod+mod)%mod);
return 0;
}