洛谷P4925 [1007]Scarlet的字元串不可能這麼可愛(計數)

題意

題目鏈接

Sol

只要知道“迴文連續子串”就能做了吧。。

想要滿足這個條件，肯定是不能出現\(aa\)或\(aba\)這種情況

如果沒有\(S\)的限制，答案為\(K * (K - 1) * \prod_{i = 3}^n (k - 2)\)

如果有\(S\)的限制就除一個\(K\)

然而考場上沒註意到會乘爆long long於是掛乘暴力分。。。。

#include<bits/stdc++.h>
#define int long long 
using namespace std;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-')f =- 1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int K, L, mod, S, W;
int mul(int a, int b) {
    return a * b % mod;
}
int fastpow(int a, int p) {
    int base = 1; a %= mod;
    while(p) {
        if(p & 1) base = 1ll * base * a % mod;
        a = 1ll * a * a % mod; p >>= 1;
    }
    return base;
}
main() {
    K = read(); L = read(); mod = read();
    S = read(); W = read();
    K %= mod;
    if(S == 0) {
        if(L == 1) cout << K;
        else if(L == 2) cout << K * (K - 1) % mod;
        else cout << 1ll * K * (K - 1) % mod * fastpow(K - 2, L - 2) % mod;
    } else {
        int base = 1;
        if(L == 1) cout << 1;
        else if(L == 2) cout << (K - 1) % mod;
        else {
            base = mul(base, K - 1);
            base = base * fastpow(K - 2, L - 2) % mod;
            cout << base;
        }
    }
}