bzoj 1008 越狱

监狱有连续编号为1...N的N个房间，每个房间关押一个犯人，有M种宗教，每个犯人可能信仰其中一种。如果相邻房间的犯人的宗教相同，就可能发生越狱，求有多少种状态可能发生越狱

1<=M<=10^8,1<=N<=10^12

水题

一共有m ^ n种状态，若没有发生越狱的情况，第一个人有m种选择，第二个人有m - 1种选择，后面每个人都有m - 1种选择，所以ans = m ^ n - m * (m - 1) ^ (n - 1)

//最后那里如果ans小于0要特判，不然会wa

#include<map>#include<cmath>#include<cstdio>#include<queue>#include<vector>#include<string>#include<cstring>#include<iostream>#include<algorithm>using namespace std;inline int read() {int x = 0, flag = 1; char ch = getchar();while (ch > '9' || ch < '0') { if (ch == '-') flag = -1; ch = getchar(); }while (ch <= '9' && ch >= '0') { x = x * 10 + ch - '0'; ch = getchar(); }return x * flag;}#define rep(ii, aa, bb) for (int ii = aa; ii <= bb; ii++)#define drp(ii, aa, bb) for (int ii = aa; ii >= bb; ii--)#define ll long long#define mod 100003ll mult(ll a, ll b) {ll c = 1, d = a % mod;while (b > 0) {if (b & 1) c = (c % mod * d % mod) % mod;b >>= 1;d = (d % mod * d % mod) % mod;}return c;}int main() {ll m, n;cin >> m >> n;int ans = (mult(m, n) - m * mult(m - 1, n - 1) % mod) % mod;if (ans < 0) ans += mod;cout << ans;return 0;}