2017 CCPC秦皇岛 B

枚举一对匹配的括号，里面和外面都是catalan数。注意到相同距离的括号对可以合并，枚举距离计一下数即可。过程中要预处理阶乘和逆元。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=2000005;
ll n, mod, b[maxn], a[maxn], A, B;
ll inv[maxn];
ll fac[maxn];
void init(){
    inv[0]=inv[1]=1;
    fac[0]=fac[1]=1;
    for(int i=2;i<=2*n;i++){
        inv[i]=inv[mod%i]*(mod-mod/i)%mod;
        fac[i]=fac[i-1]*i%mod;
    }
    for(int i=1;i<=2*n;i++){
        inv[i]=inv[i-1]*inv[i]%mod;
    }
}
ll catalan(ll n){
    if(n==0)return 1;
    ll ret=fac[2*n];
    ret=ret*inv[n+1]%mod;
    ret=ret*inv[n]%mod;
    return ret;
}
ll power(ll n, ll p){
    ll ans=1;
    ll base=n;
    while(p){
        if(p&1)ans=ans*base%mod;
        base=base*base%mod;
        p>>=1;
    }
    return ans;
}
ll Inv(ll num){
    return power(num, mod-2);
}
int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        scanf("%lld%lld%lld%lld%lld", &n, &mod, &b[0], &A, &B);
        for(ll i=1;i<=2*n;i++){
            b[i]=(A*b[i-1]+B)%mod;
            a[i]=a[i-1]+b[i]+1;
        }
        for(ll i=1;i<=2*n;i++){
            a[i]=(a[i]+a[i-1])%mod;
        }
        init();
        ll p=0;
        for(ll i=1;i<=n;i++){
            ll tmp=(a[n*2]-a[i*2-1]-a[n*2-i*2+1]+2*mod)%mod;
            tmp=tmp*catalan(i-1)%mod;
            tmp=tmp*catalan(n-i)%mod;
            p=(p+tmp)%mod;
        }
        ll q=catalan(n);
        ll ans=p*Inv(q)%mod;
        printf("%lld\n", ans);
    }
}