常数复杂度求组合数

题目链接

O(n)预处理，然后O(1)组合数：
//空间是O(n)的
//前提得是模数相同的情况下
const int maxn = 1e7;
const long long mod = 998244353;
Ll A[maxn];	//A[i]表示i的阶乘
ll B[maxn];	//B[i]表示i的阶乘关于mod的逆元
ll pow_mod(ll a,ll b)
{
    ll ret=1;
    while(b)
    {
        if(b&1) ret=ret*a%mod;
        b>>=1;
        a=a*a%mod;
    }
    return ret;
}
void Init()
{
    A[0] = 1;
    for(int i=1; i<=maxn; i++)
        A[i] = (A[i-1] * i ) % mod;
B[maxn]=pow_mod(A[maxn],mod-2);
for(int i=maxn-1;i>=0;i--)
    B[i]=B[i+1]*(i+1)%mod;
}
ll C(ll a, ll b)
{
    if(a < b)
        return 0;
    return (A[a] * B[b] %mod)*B[a-b]%mod;
}
int main()
{
Init();
    int a,b;
    while(scanf("%d%d",&a,&b)!=EOF)
    {
        cout<

常数复杂度求组合数

#include
#include
#include
using namespace std;
typedef long long LL;
const int maxn = 5e5;
const long long mod = 998244353;
LL A[maxn];
LL B[maxn];
void Init()
{
    A[0] = 1;
    for(int i=1; i<=maxn; i++){
        A[i] = (A[i-1] * i ) % mod;
    }
}

LL Ext_Gcd(LL a, LL b, LL &x, LL &y)
{
    if(b==0){
        x=1;
        y=0;
        return a;
    }
    LL d = Ext_Gcd(b, a%b, y, x);
    y-=a/b*x;
    return d;
}

LL Inv(LL a, LL n)
{
    LL x,y;
    LL d = Ext_Gcd(a,n,x,y);
    if(d == 1)
        return ((x%n)+n)%n;
    return -1;
}
LL get()
{
    for(int i=0;i>t;
    while(t--)
    {
        cin>>n>>m>>k;
        int u = k / n;
        long long sum = 0;
        for(int i = 0; i <= u; ++i)
        {
            if(k + m - 1 - i * n > 0)
            sum += C(m, i, mod) * C(k+m-1-i*n, m-1, mod) * mmp(i) ;
            sum %= mod;
            if(sum < 0)
                sum += mod;
        }
        cout<

 

貌似这个方法求C(m,n)，m和n的范围只能到5e5，太大了会超时，不过询问的时候可以常熟出结果，我也不是很懂......

#include
#include
#include
using namespace std;
typedef long long LL;
const int maxn = 5e5;
const long long mod = 998244353;
LL A[maxn];
LL B[maxn];
void Init()
{
    A[0] = 1;
    for(int i=1; i<=maxn; i++){
        A[i] = (A[i-1] * i ) % mod;
    }
}

LL Ext_Gcd(LL a, LL b, LL &x, LL &y)
{
    if(b==0){
        x=1;
        y=0;
        return a;
    }
    LL d = Ext_Gcd(b, a%b, y, x);
    y-=a/b*x;
    return d;
}

LL Inv(LL a, LL n)
{
    LL x,y;
    LL d = Ext_Gcd(a,n,x,y);
    if(d == 1)
        return ((x%n)+n)%n;
    return -1;
}
LL get()
{
    for(int i=0;i>t;
    int t1,t2;
    while(t--)
    {
        scanf("%d%d",&t1,&t2);
        cout<