/****************************** 大组合取模之:1<=n<=m<=1e6,1<=p<=1e9 使用：程序最开始调用getprime()，需要时调用C(n,m,p) 复杂度：O( n*log(n) ) ******************************/typedef long long ll;#define N 210000int PRIME[N/2];void getprime(){    bool pmark[N+1000];    memset(pmark,0,sizeof(pmark));    int pcnt=0;    PRIME[pcnt++]=2;    for(int i=3;i
     
      10^9，log(mod)*log(b),否则log(b) ***************/long long Mod_Mul(long long a,long long b,long long mod){    long long msum=0;    while(b)    {        if(b&1) msum = (msum+a)%mod;        b>>=1;        a = (a+a)%mod;    }    return msum;}long long Quk_Mul(long long a,long long b,long long mod){    bool qmflag=mod>1e9?1:0;    long long qsum=1;    while(b)    {        if(b&1) qsum = (qmflag==1) ? Mod_Mul(qsum,a,mod) : (qsum*a)%mod;        b>>=1;        a = (qmflag==1) ? Mod_Mul(a,a,mod) : (a*a)%mod;    }    return qsum;}// 得到n! 中有多少个d因子int getdn(int n,int d){    int sum=0;    while(n)    {        sum += n/d;        n/=d;    }    return sum;}ll C(int n,int m,ll p){    if(m>n) return 0;    ll sumc=1;    for(int i=0;PRIME[i]<=n;i++)    {        int cnum = getdn(n,PRIME[i])-getdn(m,PRIME[i])-getdn(n-m,PRIME[i]);        sumc = sumc*Quk_Mul(PRIME[i], cnum, p)%p;    }    return sumc;}/*int main() {    getprime();    int T;    cin>>T;    while(T--)    {        int n,m,p;        cin>>n>>m>>p;        cout<
      
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