[YDOI R1] Necklace - 洛谷

因为是方案数求和

我们考虑计算每种珠子单独贡献的方案数有

因为有二项式定理

构造

因为n不取0，便有

时间复杂度

modint + qmi code

#include <bits/stdc++.h>

#define INF (1ll<<60)
#define eps 1e-6
using namespace std;
typedef long long ll;
typedef long long LL;
typedef pair<int, int> pii;
typedef vector<int> vi;
typedef unsigned long long ull;
typedef vector<vector<int> > vii;
typedef vector<vector<vector<int> > > viii;
typedef vector<ll> vl;
typedef vector<vector<ll> > vll;
typedef vector<double> vd;
typedef vector<vector<double> > vdd;
#define time mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());//稳定随机卡牛魔 ull
const int mod=1e9+7;
template<class T>
constexpr T power(T a, LL b) {
    T res = 1;
    for (; b; b /= 2, a *= a) {
        if (b % 2) {
            res *= a;
        }
    }
    return res;
}

template<int P>
struct MInt {
    int x;
    constexpr MInt() : x{} {}
    constexpr MInt(LL x) : x{norm(x % getMod())} {}

    static int Mod;
    constexpr static int getMod() {
        if (P > 0) {
            return P;
        } else {
            return Mod;
        }
    }
    constexpr static void setMod(int Mod_) {
        Mod = Mod_;
    }
    constexpr int norm(int x) const {
        if (x < 0) {
            x += getMod();
        }
        if (x >= getMod()) {
            x -= getMod();
        }
        return x;
    }
    constexpr int val() const {
        return x;
    }
    explicit constexpr operator int() const {
        return x;
    }
    constexpr MInt operator-() const {
        MInt res;
        res.x = norm(getMod() - x);
        return res;
    }
    constexpr MInt inv() const {
        assert(x != 0);
        return power(*this, getMod() - 2);
    }
    constexpr MInt &operator*=(MInt rhs) & {
        x = 1LL * x * rhs.x % getMod();
        return *this;
    }
    constexpr MInt &operator+=(MInt rhs) & {
        x = norm(x + rhs.x);
        return *this;
    }
    constexpr MInt &operator-=(MInt rhs) & {
        x = norm(x - rhs.x);
        return *this;
    }
    constexpr MInt &operator/=(MInt rhs) & {
        return *this *= rhs.inv();
    }
    friend constexpr MInt operator*(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res *= rhs;
        return res;
    }
    friend constexpr MInt operator+(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res += rhs;
        return res;
    }
    friend constexpr MInt operator-(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res -= rhs;
        return res;
    }
    friend constexpr MInt operator/(MInt lhs, MInt rhs) {
        MInt res = lhs;
        res /= rhs;
        return res;
    }
    friend constexpr std::istream &operator>>(std::istream &is, MInt &a) {
        LL v;
        is >> v;
        a = MInt(v);
        return is;
    }
    friend constexpr std::ostream &operator<<(std::ostream &os, const MInt &a) {
        return os << a.val();
    }
    friend constexpr bool operator==(MInt lhs, MInt rhs) {
        return lhs.val() == rhs.val();
    }
    friend constexpr bool operator!=(MInt lhs, MInt rhs) {
        return lhs.val() != rhs.val();
    }
};

template<>
int MInt<0>::Mod = 1e9 + 7;

template<int V, int P>
constexpr MInt<P> CInv = MInt<P>(V).inv();

constexpr int P = 1e9 + 7;
using Z = MInt<P>;

const int N=2e5+9;

Z inv[N],f[N],invf[N];

//组合数
void init(){
    inv[1]=1;
    for(int i=2;i<N;i++){
        inv[i]=inv[mod%i]*(mod-mod/i);
    }
    f[0]=1,invf[0]=1;
    for(int i=1;i<N;i++){
        f[i]=f[i-1]*i;
        invf[i]=invf[i-1]*inv[i];
    }
}
Z C(ll n,ll m){
    if(n<m || m<0){
        return 0;
    }
    return f[n]*invf[m]*invf[n-m];
}
Z qmi(Z a,ll b){
    Z res=1;
    while(b){
        if(b&1){
            res=res*a;
        }
        b>>=1;
        a=a*a;
    }
    return res;
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    int n;
    cin>>n;
    vi a(n+1);
    ll sum=0;
    for(int i=1;i<=n;i++){
        cin>>a[i];
        sum+=a[i];
    }
    vi v(n+1);
    for(int i=1;i<=n;i++){
        cin>>v[i];
    }
    Z ans=0;
    for(int i=1;i<=n;i++){
        ans+=qmi(2,sum-a[i])*(qmi(v[i]+1,a[i])-1);
    }
    cout<<ans<<'\n';
    return 0;
}