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
1 /* 2 * Problem: BZOJ 1008 3 * Author: SHJWUDP 4 * Created Time: 2015/4/28 星期二 16:29:02 5 * File Name: 233.cpp 6 * State: Accepted 7 * Memo: 容斥 8 */ 9 #include <iostream> 10 #include <cstdio> 11 #include <cstring> 12 #include <algorithm> 13 14 using namespace std; 15 16 typedef long long int64; 17 18 const int MOD=1e5+3; 19 20 int64 M, N; 21 int64 pow(int64 a, int64 n, int64 mod) { 22 int64 res=1; 23 while(n) { 24 if(n&1) res=(res*a)%mod; 25 a=(a*a)%mod; 26 n>>=1; 27 } 28 return res; 29 } 30 int main() { 31 #ifndef ONLINE_JUDGE 32 freopen("in", "r", stdin); 33 //freopen("out", "w", stdout); 34 #endif 35 while(~scanf("%lld%lld", &M, &N)) { 36 int64 ans=M*(pow(M, N-1, MOD)-pow(M-1, N-1, MOD))%MOD; 37 printf("%lld\n", (ans+MOD)%MOD); 38 } 39 return 0; 40 }
