求组合数C（m,n）的多种计算方法

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1000;
int C[MAXN+1][MAXN+1];
//求排列组合数C(m,n)  上面为m,下面为n  m<n
//C(m,n)=n!/m!/(n-m)!=n*(n-1)*..*(n-m+1)/m!.
int baoli_C(int m,int n)  //暴力法这里n<=15
{
    int summ=1,sumn=1;
    //其实算C(m,n)只要计算min(m,n-m)次就可以了
    if(m>n-m)
        m=n-m;
    for(int i=1;i<=m;i++){
        summ*=i;
        sumn=sumn*(n-i+1);
    }
    return sumn/summ;
}

void dabiao_C(){    //打表，数据为int,注意溢出数据   n<60
    //C(n, m)  = C(n -1, m - 1) + C(n - 1, m)
     for(int i=0;i<MAXN;i++)
    {
//         C[i][0]=1; C[0][i]=0;//该写法顺序是错误的，因为这样写C[0][0]=0;
         C[0][i]=0;C[i][0]=1;
    }
     for(int i=1;i<MAXN;i++)
        for(int j=1;j<MAXN;j++)
        C[i][j]=C[i-1][j-1]+C[i-1][j];
}

int main(){
    dabiao_C();
    int m,n;
    while(cin>>m>>n){  //m<n
        cout<<C[m][n]<<endl;
    }
}

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int MAXN = 1000;
#define mod 998244353
const int maxn = 1e5+10;
ll fac[maxn];//maxn应该小于1e5,这样Lucas定理才适用
void Init(){  //对阶乘打表
    fac[0]=1;
    for(int i=1;i<=maxn-5;i++){
        fac[i]=fac[i-1]*i%mod;
    }
}

ll quickpow(ll a,ll b){
     ll t=a,ans=1;
     while(b!=0){
        if(b&1==1)
            ans=(ans*t)%mod;
        t=t*t%mod;
        b>>=1;
     }
     return ans%mod;
}

ll C(ll n,ll m){
    if(m>n)
        return 0;
     return fac[n]*quickpow(fac[m]*fac[n-m],mod-2)%mod;
}

ll Lucas(ll n,ll m){
    if(m==0)
        return 1;
    return Lucas(n/mod,m/mod)*C(n%mod,m%mod)%mod;
}
int main(){
    ll n,m;
    Init();
    while(cin>>n>>m){
        cout<<C(n,m)<<endl;
    }
}

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int MAXN = 1000;
#define mod 998244353
ll pow(ll a, ll b, ll m)
{
    ll ans = 1;
    a %= m;
    while(b)
    {
        if(b & 1)ans = (ans % m) * (a % m) % m;
        b /= 2;
        a = (a % m) * (a % m) % m;
    }
    ans %= m;
    return ans;
}
ll inv(ll x, ll p)//x关于p的逆元，p为素数
{
    return pow(x, p - 2, p);
}
ll C(ll n, ll m, ll p)//组合数C(n, m) % p = (n!/m!/(n-m)!)%mod = n!*inv(m!*(n-m)!)%mod = n!*(m!*((n-m)!)^(mod-2))%mod
{
    if(m > n)return 0;
    ll up = 1, down = 1;//分子分母;
    for(int i = n - m + 1; i <= n; i++)up = up * i % p;
    for(int i = 1; i <= m; i++)down = down * i % p;
    return up * inv(down, p) % p;
}
ll Lucas(ll n, ll m, ll p)
{
    if(m == 0)return 1;
    return C(n % p, m % p, p) * Lucas(n / p, m / p, p) % p;
}
int main(){
    int m,n;
    while(cin>>m>>n){  //m<n
        cout<<Lucas(m,n,mod)<<endl;
    }
}

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e6 + 10;
const int mod = 1e9 + 7;
ll pow(ll a, ll b, ll m)
{
    ll ans = 1;
    a %= m;
    while(b)
    {
        if(b & 1)ans = (ans % m) * (a % m) % m;
        b /= 2;
        a = (a % m) * (a % m) % m;
    }
    ans %= m;
    return ans;
}
ll extgcd(ll a, ll b, ll& x, ll& y)
//求解ax+by=gcd(a, b)
//返回值为gcd(a, b)
{
    ll d = a;
    if(b)
    {
        d = extgcd(b, a % b, y, x);
        y -= (a / b) * x;
    }
    else x = 1, y = 0;
    return d;
}
ll mod_inverse(ll a, ll m)
//求解a关于模上m的逆元
//返回-1表示逆元不存在
{
    ll x, y;
    ll d = extgcd(a, m, x, y);
    return d == 1 ? (m + x % m) % m : -1;
}

ll Mul(ll n, ll pi, ll pk)//计算n! mod pk的部分值  pk为pi的ki次方
//算出的答案不包括pi的幂的那一部分
{
    if(!n)return 1;
    ll ans = 1;
    if(n / pk)
    {
        for(ll i = 2; i <= pk; i++) //求出循环节乘积
            if(i % pi)ans = ans * i % pk;
        ans = pow(ans, n / pk, pk); //循环节次数为n / pk
    }
    for(ll i = 2; i <= n % pk; i++)
        if(i % pi)ans = ans * i % pk;
    return ans * Mul(n / pi, pi, pk) % pk;//递归求解
}

ll C(ll n, ll m, ll p, ll pi, ll pk)//计算组合数C(n, m) mod pk的值 pk为pi的ki次方
{
    if(m > n)return 0;
    ll a = Mul(n, pi, pk), b = Mul(m, pi, pk), c = Mul(n - m, pi, pk);
    ll k = 0, ans;//k为pi的幂值
    for(ll i = n; i; i /= pi)k += i / pi;
    for(ll i = m; i; i /= pi)k -= i / pi;
    for(ll i = n - m; i; i /= pi)k -= i / pi;
    ans = a * mod_inverse(b, pk) % pk * mod_inverse(c, pk) % pk * pow(pi, k, pk) % pk;//ans就是n! mod pk的值
    ans = ans * (p / pk) % p * mod_inverse(p / pk, pk) % p;//此时用剩余定理合并解
    return ans;
}

ll Lucas(ll n, ll m, ll p)
{
    ll x = p;
    ll ans = 0;
    for(ll i = 2; i <= p; i++)
    {
        if(x % i == 0)
        {
            ll pk = 1;
            while(x % i == 0)pk *= i, x /= i;
            ans = (ans + C(n, m, p, i, pk)) % p;
        }
    }
    return ans;
}

int main()
{
    ll n, m, p;
    while(cin >> n >> m >> p)
    {
        cout<<Lucas(n, m, p)<<endl;
    }
    return 0;
}

#include<bits/stdc++.h>
using namespace std;
#define mod 998244353
#define ll long long
ll fac[100005];
void Init(){
fac[0]=1;
for(int i=1;i<=100000;i++)
fac[i]=(fac[i-1]*i)%mod;
}

long long quickmod(int x,int y)
{
if(x==0)
return 0;
long long t=x,ans=1;
while(y!=0)
{
if(y&1==1)
ans=(ans*t)%mod;
t=(t*t)%mod;
y>>=1;
}
return ans;
}

ll C(int n,int m){
return fac[n]/fac[n-m]*quickmod(fac[m],mod-2)%mod;
}

ll Lucas(int n,int m){
if(m==0)
return 1;
return Lucas(n/mod,m/mod)*C(n%mod,m%mod)%mod;
}

int main()
{
Init();
int T;
scanf("%d",&T);
int a,b,n,m;
while(T--){
scanf("%d%d%d%d",&a,&b,&n,&m);
long long h,ans;
h=Lucas(n-1,m-1);
// cout<<quickmod(a,n-m)%mod<<endl;
// cout<<quickmod(b,m-1)%mod<<endl;
ans=quickmod(a,n-m)*quickmod(b,m-1)%mod*h;
cout<<ans%mod<<endl;
}
}

/*
10
2 0 5 3
2 3 3 5
5 1 4 6
5 1 6 4
0 2 6 4
2 0 6 4
*/