# BZOJ 1008 越狱

Problem 1008. -- [HNOI2008]越狱

## 1008: [HNOI2008]越狱

Time Limit: 1 Sec  Memory Limit: 162 MB
Submit: 9540  Solved: 4128

## Description

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

## Input

输入两个整数M，N.1<=M<=10^8,1<=N<=10^12

## Output

可能越狱的状态数，模100003取余

2 3

6

## HINT

6种状态为(000)(001)(011)(100)(110)(111)

AC代码：

# include <iostream>
# include <cstring>
# include <cstdlib>
# include <cstdio>
# include <cmath>
# include <map>
# include <queue>
# include <stack>
# include <vector>
# include <fstream>
# include <algorithm>
using namespace std;
# define bug puts("")
# define pi acos(-1.0)
# define mem(a,b) memset(a,b,sizeof(a))
# define IOS ios::sync_with_stdio(false)
# define FO(i,n,a) for(int i=n; i>=n; --i)
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
typedef unsigned long long ULL;
typedef long long LL;
inline int Scan() {
int x=0,f=1; char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
///coding...................................

LL mod_pow(LL x,LL n,LL mod) {
LL res=1;
while(n>0) {
if(n&1)res=res*x%mod;
x=x*x%mod;
n>>=1;
}
return res;
}

int main()
{
#ifdef FLAG
freopen("in.txt","r",stdin);
#endif /// FLAG
LL m,n,ans,mod=100003;
while(~scanf("%lld%lld",&m,&n)) {
LL t1=mod_pow(m,n,mod);
LL t2=m*mod_pow(m-1,n-1,mod);
ans=((t1-t2)%mod+mod)%mod;
cout<<ans<<endl;
}
return 0;
}

posted @ 2017-07-31 10:27  韵祈  阅读(86)  评论(0编辑  收藏