[HNOI2008]越狱

题面

 

一道很水的数论题。

考虑不发生越狱的情况：第一个人可信m种宗教，第二个人只能信m - 1种宗教，第三个人也只能信m - 1种宗教，那么这割状态总数就是m * (m - 1)^{n - 1}。

题中说求越狱状态总数，那么就是mⁿ - m * (m - 1)^{n - 1}，别忘取模。

 1 #include<cstdio>
 2 #include<iostream>
 3 #include<cmath>
 4 #include<algorithm>
 5 #include<cstring>
 6 #include<cstdlib>
 7 #include<queue>
 8 #include<stack>
 9 #include<vector>
10 #include<cctype>
11 using namespace std;
12 #define enter puts("")
13 #define space putchar(' ')
14 #define Mem(a) memset(a, 0, sizeof(a))
15 typedef long long ll;
16 typedef double db;
17 const int INF = 0x3f3f3f3f;
18 const db eps = 1e-8;
19 const ll mod = 100003;
20 //const int maxn = ;
21 inline ll read()
22 {
23     ll ans = 0;
24     char ch = getchar(), last = ' ';
25     while(!isdigit(ch)) {last = ch; ch = getchar();}
26     while(isdigit(ch))
27     {
28         ans = ans * 10 + ch - '0'; ch = getchar();
29     }
30     if(last == '-') ans = -ans;
31     return ans;
32 }
33 inline void write(ll x)
34 {
35     if(x < 0) putchar('-'), x = -x;
36     if(x >= 10) write(x / 10);
37     putchar(x % 10 + '0');
38 }
39 
40 ll n, m;
41 
42 ll quickpow(ll a, ll b)
43 {
44     a %= mod;
45     ll ret = 1;
46     while(b)
47     {
48         if(b & 1) ret = ret * a % mod;
49         a = a * a % mod; b >>= 1;
50     }
51     return ret;
52 }
53 
54 int main()
55 {
56     m = read(); n = read();
57     write((quickpow(m, n) - m * quickpow(m - 1, n - 1) % mod + mod) % mod); enter;
58 }

View Code

 

标签：10,ch,ll,越狱,ans,HNOI2008,include,mod
来源： https://blog.51cto.com/u_15234622/2830848