参考：http://blog.csdn.net/u014610830/article/details/49493279
这道题做起来感觉非常奇怪啊……头一次见把mu推出来再推没了的……
[sum_{i=a}^{b}lcm(i,b)
]
[sum_{i=a}^{b}frac{i*b}{gcd(i,b)}
]
[sum_{d|b}sum_{i=a}^{b}[gcd(i,b)==d]frac{ib}{d}
]
[sum_{d|b}sum_{i=left lfloor frac{a}{d}
ight
floor}^{left lfloor frac{b}{d}
ight
floor}[gcd(i,b)==1]frac{idb}{d}
]
[bsum_{d|b}sum_{i=left lfloor frac{a}{d}
ight
floor}^{left lfloor frac{b}{d}
ight
floor}[gcd(i,b)==1]i
]
[bsum_{d|b}sum_{i=left lfloor frac{a}{d}
ight
floor}^{left lfloor frac{b}{d}
ight
floor}isum_{k|gcd(i,frac{b}{d})}mu(k)
]
[bsum_{d|b}sum_{k|frac{b}{d}}mu(k)sum_{i=left lfloor frac{a}{d}
ight
floor}^{left lfloor frac{b}{d}
ight
floor,k|i}i
]
[bsum_{d|b}sum_{k|frac{b}{d}}mu(k)ksum_{i=left lfloor frac{a}{dk}
ight
floor}^{left lfloor frac{b}{dk}
ight
floor}i
]
设
[f(a,b)=sum_{i=a}^{b}i
]
[bsum_{d|b}sum_{k|frac{b}{d}}mu(k)kf(left lfloor frac{a}{dk}
ight
floor,left lfloor frac{b}{dk}
ight
floor)
]
枚举dk
[bsum_{d|b}f(frac{a}{d},frac{b}{d})sum_{k|d}mu(k)k
]
其中f可以( O(1) )算，设( g(x)=sum_{k|x}mu(k)k )，容易发现这是积性的，显然设p为质数，( g(p)=1-p,g(p^k)=1-p )，于是这样就把mu推掉了
时间复杂度很不科学但是还是能过
#include<iostream>
#include<cstdio>
#include<vector>
using namespace std;
const int N=100005,m=100000,inv2=500000004,mod=1e9+7;
int T,n,q[N],tot,a,b,p[N],cnt,len[N];
bool v[N];
vector<int>fac;
void dfs(int re,int l)
{
if(l>cnt)
{
fac.push_back(re);
return;
}
int tmp=1;
dfs(re,l+1);
for(int i=1;i<=len[l];i++)
{
tmp*=p[l];
dfs(re*tmp,l+1);
}
}
int main()
{
v[1]=1;
for(int i=2;i<=m;i++)
{
if(!v[i])
q[++tot]=i;
for(int j=1;j<=tot&&i*q[j]<=m;j++)
{
int k=i*q[j];
v[k]=1;
if(i%q[j]==0)
break;
}
}
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&a,&b);
cnt=0;
int now=b;
for(int i=1;i<=tot&&q[i]*q[i]<=now;i++)
{
if(now%q[i]==0)
{
p[++cnt]=q[i];
len[cnt]=0;
}
while(now%q[i]==0)
{
now/=q[i];
len[cnt]++;
}
}
if(now>1)
{
p[++cnt]=now;
len[cnt]=1;
}
fac.clear();
dfs(1,1);//cout<<cnt<<endl;
long long ans=0ll;
for(int i=0;i<fac.size();i++)
{
int v=fac[i];
long long tt=a+v-1,tmp,re=1ll;
tmp=(((tt/v+b/v)%mod)*((b/v-tt/v+1+mod)%mod)%mod)*inv2%mod;
for(int j=1;j<=cnt;j++)
if(v%p[j]==0)
re=re*((1-p[j]+mod)%mod)%mod;
ans=((ans+re*tmp%mod)%mod+mod)%mod;
}
printf("%lld
",ans*b%mod);
}
return 0;
}