分析:展开欧拉函数化简得到,我们要处理的就是gcd(a,b)=igcd(a,b)=i时的G(a,b)值G(a,b)值,以及有多少对a,b的gcd为i,G(a,b)=ϕ(i)/i,G(a,b)=ϕ(i)/i,预处理欧拉函数和逆元即可。对于一个数i,在a∈[1,n],b∈[1,m]的范围内,设f[i]为gcd为(i,2i,3i...)的对数,显然 : f[i]=[n/i]∗[m/i],那么我们从大到小维护f[i],因为我们要的是 gcd=i的对数,所以要把 gcd=i的倍数的情况减去。(d枚举d所有小于max(n,m)的倍数,再减去d的倍数方案数)。
代码:
#pragma comment(linker, "/STACK:102400000,102400000")///手动扩栈
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cassert>
#include<string>
#include<cstdio>
#include<bitset>
#include<vector>
#include<cmath>
#include<ctime>
#include<stack>
#include<queue>
#include<deque>
#include<list>
#include<set>
#include<map>
using namespace std;
#define debug test
#define mst(ss,b) memset((ss),(b),sizeof(ss))
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define all(x) (x).begin(),(x).end()
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define ll long long
#define ull unsigned long long
#define pb push_back
#define mp make_pair
#define inf 0x3f3f3f3f
#define eps 1e-10
#define PI acos(-1.0)
typedef pair<int,int> PII;
const ll mod = 1e9+7;
const int N = 1e6+10;
ll t,n,m,p;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qp(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
int to[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
bool vis[N];
ll phi[N],prim[N/10],cnt;
void gephi(){///线性筛欧拉函数
phi[1] = 1;
for(int i=2;i<N;i++){
if(!vis[i]) {
phi[i] = i-1;
prim[++cnt] = i;
}
for(int j=1;j<=cnt;j++){
int tp = prim[j];
if(i*tp>N) break;
vis[i*tp]=true;
if(i%tp==0) {
phi[i*tp]=phi[i]*tp;
break;
}else phi[i*tp]=phi[i]*phi[tp];
}
}
}
ll inv[N],gd[N];
int main() {
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
gephi();
inv[1]=1;
cin>>t;
while(t--) {
ll ans(0);
cin>>n>>m>>p;
if(n>m) swap(n,m);
rep(i,2,n) inv[i]=(p-p/i)*inv[p%i]%p;
rep(i,1,n) gd[i]=(n/i)*(m/i)%p;
for(int i=n/2;i>=1;i--) {///注意要从后往前递推
for(int j=2*i;j<=n;j+=i) {
gd[i]-=gd[j];
if(gd[i]<0) gd[i]+=p;
}
}
for(int i=1;i<=n;i++)
(ans += gd[i]*i%p*inv[phi[i]]%p )%=p;
cout<<ans<<endl;
}
return 0;
}