越狱 BZOJ

题目传送门

思路：这个题正着想不容易想，但是我们可以去想怎么才能让越狱不发生，也就是任何两个相邻的人的信仰都不相同，然后再用所有的情况减去这个情况就好了。

#include <algorithm>#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <list>#include <map>#include <queue>#include <set>#include <stack>#include <string>#include <vector>#define MAXN 1000010#define MAXE 5#define INF 0x7ffffff#define MOD 100003#define LL long long#define ULL unsigned long long#define pi 3.14159using namespace std;LL MI(LL a, LL b) {    LL ans = 1;    while (b) {        if (b & 1) {            ans = ans * a % MOD;        }        b >>= 1;        a = a * a % MOD;    }    return ans % MOD;}int main() {    std::ios::sync_with_stdio(false);    LL m, n;    cin >> m >> n;    LL sum1 = MI(m, n);    LL sum2 = m;    sum2 %= MOD;    sum2 = sum2 * MI(m - 1, n - 1) % MOD;    cout << (sum1 - sum2 + MOD) % MOD << endl;    return 0;}