bzoj 1008: [HNOI2008]越狱

题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=1008

简单容斥：

总状态数 A=M^N

不越狱的状态数 B=M*(M-1)^(N-1)

越狱的状态数 ans=A-B

/*
 * Problem: BZOJ 1008 
 * Author:  SHJWUDP
 * Created Time:  2015/4/28 星期二 16:29:02
 * File Name: 233.cpp
 * State: Accepted
 * Memo: 容斥
 */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long int64;

const int MOD=1e5+3;

int64 M, N;
int64 pow(int64 a, int64 n, int64 mod) {
    int64 res=1;
    while(n) {
        if(n&1) res=(res*a)%mod;
        a=(a*a)%mod;
        n>>=1;
    }
    return res;
}
int main() {
#ifndef ONLINE_JUDGE
    freopen("in", "r", stdin);
    //freopen("out", "w", stdout);
#endif
    while(~scanf("%lld%lld", &M, &N)) {
        int64 ans=M*(pow(M, N-1, MOD)-pow(M-1, N-1, MOD))%MOD;
        printf("%lld
", (ans+MOD)%MOD);
    }
    return 0;
}