EOJ Monthly 2021.9 Sponsored by TuSimple(A)

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EOJ Monthly 2021.9 Sponsored by TuSimple

A.Amazing.Discovery

题意:

给出 a , b , n a,b,n abn,求
S = ( a + b ) n + ( a − b ) n ( m o d   p ) S=(a+\\sqrt b)^n+(a-\\sqrt b)^n(mod~p) S=(a+b )n+(ab )n(mod p)

思路:

因为 b \\sqrt b b 可能在模 m o d mod mod的情况下没有整数解。

解法1:

S n = ( a + b ) n + ( a − b ) n S_n=(a+\\sqrt b)^n+(a-\\sqrt b)^n Sn=(a+b )n+(ab )n

S 2 n = ( a + b ) 2 n + ( a − b ) 2 n = S n 2 − 2 ( a + b ) n ( a − b ) n = S n 2 − 2 ( a 2 − b ) n S_{2n}=(a+\\sqrt b)^{2n}+(a-\\sqrt b)^{2n}=S_n^2-2(a+\\sqrt b)^n(a-\\sqrt b)^n=S_n^2-2(a^2-b)^n S2n=(a+b )2n+(ab )2n=Sn22(a+b )n(ab )n=Sn22(a2b)n

S 2 n + 1 = ( a + b ) 2 n + 1 + ( a − b ) 2 n + 1 = S n S n + 1 − ( a + b ) n ( a − b ) n + 1 − ( a + b ) n + 1 ( a − b ) n = S n 2 − 2 a ( a 2 − b ) n S_{2n+1}=(a+\\sqrt b)^{2n+1}+(a-\\sqrt b)^{2n+1}=S_nS_{n+1}-(a+\\sqrt b)^n(a-\\sqrt b)^{n+1}-(a+\\sqrt b)^{n+1}(a-\\sqrt b)^n=S_n^2-2a(a^2-b)^n S2n+1=(a+b )2n+1+(ab )2n+1=SnSn+1(a+b )n(ab )n+1(a+b )n+1(ab )n=Sn22a(a2b)n

分奇数和偶数往下进行分治(记忆化一下,减少算的次数)

#include <bits/stdc++.h>
#define lson rt << 1
#define rson (rt << 1) | 1

using namespace std;

typedef long long ll;
const int mod = 998244353;
ll qpow(ll x, ll y) {
    ll ans = 1;
    while(y) {
        if(y & 1) ans = ans * x % mod;
        x = x * x % mod;
        y >>= 1;
    }
    return ans;
}
map<ll, ll> mp;
ll solve(ll a, ll b, ll n) {
    if(mp.find(n) != mp.end()) return mp[n];
    else if(n == 1) return mp[1] = 2ll*a;
    else {
        if(n % 2 == 0) {
            ll tmp = solve(a, b, n/2);
            return mp[n] = ((tmp * tmp % mod - 2ll * qpow((a*a%mod-b+mod)%mod, n/2) % mod) % mod + mod) % mod;
        }
        else {
            int k = n/2;
            ll tmp1 = solve(a, b, k), tmp2 = solve(a, b, n-k);
            return mp[n] = ((tmp1 * tmp2 % mod - 2ll * a % mod * qpow((a*a%mod-b+mod)%mod, n/2) % mod) % mod + mod) % mod;
        }
    }
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    freopen("out.txt", "w", stdout);
#endif
    ll a, b, n;
    cin >> a >> b >> n;
    cout << solve(a, b, n); 

}

解法2:

二次剩余(简单理解)

用二次剩余 C i p o l l a Cipolla Cipolla算法中的类似复数域的东西可以直接求解。

#include <bits/stdc++.h>
using namespace<

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