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题意:求C(m+n-2,m-1)%p,其中1<=m,n,p<=1e5,p不一定是素数
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <map>
//#define IO ios::sync_with_stdio(false),cin.tie(0), cout.tie(0);
//#pragma comment(linker, "/STACK:1024000000,1024000000")
void ex_gcd(int a, int b, int &d, int &x, int &y) { if (!b) { x = 1; y = 0; d = a; } else { ex_gcd(b, a%b, d, y, x); y -= x * (a / b); }; }
int gcd(int a, int b) { return b ? gcd(b, a%b) : a; }
int lcm(int a,int b){return a/gcd(a,b)*b;}//Ïȳýºó³Ë·ÀÒç³ö
int inv_exgcd(int a, int m) { int d, x, y;ex_gcd(a, m, d, x, y);return d == 1 ? (x + m) % m : -1; }
typedef long long ll;
const int maxn=1e5;
using namespace std;
ll cnt=0;
int a[maxn];
int primer[maxn];
void isprime()
{
cnt=0;
for(int i=2;i<maxn;++i)
{
if(!a[i])
{
primer[cnt++]=i;
for(int j=i*2;j<maxn;j+=i)
a[j]=1;
}
}
}
ll Pow(ll a,ll n,ll mod)
{
ll ans=1;
while(n)
{
if(n&1)
ans=ans*a%mod;
a=a*a%mod;
n>>=1;
}
return ans;
}
ll C(ll n,ll p,ll pk)
{
if(n==0)return 1;
ll ans=1;
for(ll i=2;i<=pk;++i)
if(i%p)ans=ans*i%pk;
ans=Pow(ans,n/pk,pk);
for(ll k=n%pk,i=2;i<=k;++i)
if(i%p)ans=ans*i%pk;
return ans*C(n/p,p,pk)%pk;
}
ll reverse(ll a,ll m)
{
if(!a)return 0;
ll y=0,x=1,r=a%m,q,t,M=m;
if(r<0)r+=m;
while(m%r)
{
a=m,m=r,q=a/m,r=a%m;
t=x,x=y-x*q,y=t;
}
if(r!=1)return 0;
if(x<0)x+=M;
return x;
}
ll ex_lucas(ll n,ll m,ll p,ll pi,ll pk)
{
ll i,j,k=0,a,b,c,ans;
a=C(n,pi,pk),b=C(m,pi,pk),c=C(n-m,pi,pk);
for(i=n;i;i/=pi)k+=i/pi;
for(i=m;i;i/=pi)k-=i/pi;
for(i=n-m;i;i/=pi)k-=i/pi;
ans=a*reverse(b,pk)%pk*reverse(c,pk)%pk*Pow(pi,k,pk)%pk;
return ans*(p/pk)%p*reverse(p/pk,pk)%p;
}
ll solve(ll n,ll m,ll x)
{
ll ans=0,p=x;
isprime();
for(ll i=0;primer[i]*primer[i]<=x;++i)
{
if(x%primer[i]==0)
{
ll pk=1;
while(x%primer[i]==0)
x/=primer[i],pk*=primer[i];
ans=(ans+ex_lucas(n,m,p,primer[i],pk))%p;
}
}
if(x>1)
ans=(ans+ex_lucas(n,m,p,x,x))%p;
return ans;
}
int main()
{
ll n,m,p;
ll t;
scanf("%lld",&t);
while(t--)
{
cin>>n>>m>>p; n+=m-2,m--;
ll ans=solve(n,m,p);
printf("%lld\n",ans);
}
return 0;
}