BZOJ 1008 [HNOI2008]越狱 (简单排列组合 + 快速幂)

1008: [HNOI2008]越狱

Time Limit: 1 Sec Memory Limit: 162 MB
Submit: 10503 Solved: 4558
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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
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

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

Sample Input

2 3

Sample Output

HINT

　　6种状态为(000)(001)(011)(100)(110)(111)

Source

析：很容易知道，一共有 m^n 种方案，然后我们可以求全部都不一样的，也就是m*(m-1)*(m-1)..*(m-1) 也就是m * (m-1)^(n-1)。由于 m 和 n 比较大，所以要用快速幂。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#include <list>

#include <assert.h>

#include <bitset>

#include <numeric>

#define debug() puts("++++")

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define fi first

#define se second

#define pb push_back

#define sqr(x) ((x)*(x))

#define ms(a,b) memset(a, b, sizeof a)

#define sz size()

#define pu push_up

#define pd push_down

#define cl clear()

#define all 1,n,1

#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e17;

const double inf = 1e20;

const double PI = acos(-1.0);

const double eps = 1e-3;

const int maxn = 2e5 + 10;

const int maxm = 3e5 + 10;

const int mod = 100003;

const int dr[] = {-1, 0, 1, 0};

const int dc[] = {0, -1, 0, 1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c) {

  return r >= 0 && r < n && c >= 0 && c < m;

}

LL fast_pow(LL a, LL n){

  LL res = 1;

  a %= mod;

  while(n){

    if(n&1)  res = res * a % mod;

    n >>= 1;

    a = a * a % mod;

  }

  return res;

}

int main(){

  LL n, m;

  scanf("%lld %lld", &m, &n);

  LL ans = fast_pow(m, n) - m * fast_pow(m-1, n-1) % mod;

  ans = (ans % mod + mod) % mod;

  printf("%lld\n", ans);

  return 0;

}

巴特西