51Nod-1190 最小公倍数之和V2
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#include<bits/stdc++.h>
#define me(a,x) memset(a,x,sizeof(a))
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);
#define sc scanf
#define itn int
#define STR clock_t startTime = clock();
#define END clock_t endTime = clock();cout << double(endTime - startTime) / CLOCKS_PER_SEC *1000<< "ms" << endl;
using namespace std;
const int N=2e6+5;
const long long mod=1e9+7;
const int inv2=mod+1>>1;
const long long mod2=998244353;
const int oo=0x7fffffff;
const int sup=0x80000000;
typedef long long ll;
typedef unsigned long long ull;
template <typename it>void db(it *begin,it *end){while(begin!=end)cout<<(*begin++)<<" ";puts("");}
template <typename it>
string to_str(it n){string s="";while(n)s+=n%10+'0',n/=10;reverse(s.begin(),s.end());return s;}
template <typename it>int o(it a){cout<<a<<endl;return 0;}
inline ll mul(ll a,ll b,ll c){ll ans=0;for(;b;b>>=1,a=(a+a)%c)if(b&1)ans=(ans+a)%c;return ans;}
inline ll ksm(ll a,ll b,ll c){ll ans=1;for(;b;b>>=1,a=mul(a,a,c))if(b&1)ans=mul(ans,a,c);return ans;}
inline void exgcd(ll a,ll b,ll &x,ll &y){if(!b)x=1,y=0;else exgcd(b,a%b,y,x),y-=x*(a/b);}
int prime[N],tot=0;
bool vis[N];
int f[N];
void pre(){
f[1]=1;
for(int i=2;i<N;i++){
if(!vis[i])prime[++tot]=i,f[i]=1-i;
for(int j=1;j<=tot&&i*prime[j]<N;j++){
vis[i*prime[j]]=1;
if(i%prime[j]==0){
f[i*prime[j]]=f[i];
break;
}else f[i*prime[j]]=f[i]*f[prime[j]];
}
}
}
vector<int>dis;
int cnt;
int p[N],len[N];
void dfs(int res,int pos){
if(pos>=cnt){
dis.push_back(res);
return;
}
int x=1;
dfs(res,pos+1);
for(int i=0;i<len[pos];i++){
x*=p[pos];
dfs(res*x,pos+1);
}
}
ll cal(int m,int n){
int sz=dis.size();
ll ans=0;
for(int x=0;x<sz;x++){
int T=dis[x];
if(T<N){
ans+=1LL*(m/T)*(m/T+1)/2%mod*f[T];
ans=(ans+mod)%mod;
}
else{
int k=T;
int res=1;
for(int i=1;i<=tot&&1LL*prime[i]*prime[i]<=k;i++){
if(k%prime[i]==0){
res*=1-prime[i];
while(k%prime[i]==0)k/=prime[i];
}
}
if(k>1)res*=1-k;
ans+=1LL*(m/T)*(m/T+1)/2%mod*res;
ans=(ans+mod)%mod;
}
}
return ans*n%mod;
}
ll qp(ll a,ll b){
ll ans=1;
for(;b;b>>=1,a=a*a)if(b&1)ans=ans*a;
return ans;
}
int main(){
//STR IN
pre();
int t;cin>>t;
while(t--){
int m,n;
sc("%d%d",&m,&n);
ll ans=1,k=n;
cnt=0;
dis.clear();
for(int i=1;i<=tot&&1LL*prime[i]*prime[i]<=n;i++){
if(n%prime[i]==0){
p[cnt++]=prime[i];
len[cnt-1]=0;
int k=0;
while(n%prime[i]==0)n/=prime[i],k++,len[cnt-1]++;
ll res=1;
ll x=qp(prime[i],k);
x=(x+1)*(x-1)/(prime[i]+1)%mod;
x=x*prime[i]%mod;
res+=x;
if(res>=mod)res%=mod;
ans=ans*res;
if(ans>=mod)ans%=mod;
}
}
if(n>1)ans=ans*((1LL*n*n-n+1)%mod)%mod,p[cnt++]=n,len[cnt-1]=1;
dfs(1,0);
ans=(ans+1)*k%mod*inv2%mod;
ans-=cal(m,k);
ans+=1LL*m*k/__gcd(1LL*m,k);
printf("%lld\n",(ans+mod)%mod);
}
// END
}