codeforces magic five －－快速幂模

题目链接：http://codeforces.com/contest/327/problem/C

首先先算出一个周期里面的值，保存在ans里面，就是平常的快速幂模m做法．

然后要计算一个公式，比如有k个部分，那么对于没一个位置i, 都有2^i + 2^(i+n) + ... + 2^(i+(k-1)*n) = 2^i(1 + 2^n + ... + 2^((k-1)*n)) = 2^i * (1-2^(n*k))/(1-2^n)

所以结果就是ans * (1-2^(n*k))/(1-2^n) % MOD;

然后就是关键计算(1-2^(n*k))/(1-2^n) % MOD；

用到费马小定理a^(p-1)同余于1(mod 1)．p是一个质数，那么a^(p-2) * a 同余于１(mod 1) ，所以a  的逆元就是 a^(p-2)

MOD是一个质数,所以(1-2^(n*k))/(1-2^n) % MOD = (2^(n*k)-1)/(2^n-1) % MOD = (2^(n*k)-1)%MOD * ((2^n-1)^(MOD-2))%MOD

 /*
  * =====================================================================================
  *       Filename:  magic.cpp
  *        Created:  19/07/2013 12:27:18
  *         Author:  liuxueyang (lxy), 1459917536@qq.com
  *   Organization:  Hunan University
  *
  * =====================================================================================
  */

 /*
 ID: zypz4571
 LANG: C++
 TASK: magic
 */
 #include <cstdlib>
 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <cctype>
 #include <algorithm>
 #include <queue>
 #include <set>
 #include <stack>
 #include <map>
 #include <list>
 using namespace std;
 #define INF 0x3f3f3f3f
 const double eps=1e-;
 char a[];
 const int MOD=;
 #define LL long long
 int k;
 LL quick(LL a, LL b) {
     LL ans=;
     while (b) {
         if(b&) ans=(ans*a)%MOD; b/=; a*=a; a%=MOD;
     }
     return ans;
 }
 int main ( int argc, char *argv[] )
 {
 #ifndef ONLINE_JUDGE
     freopen("in.txt", "r", stdin);
 #endif
     int k; string s; cin>>s>>k; LL n=s.size();
     LL a=quick(,n*k)-+MOD; a=(a+MOD)%MOD;
     LL b=quick((quick(,n)-+MOD)%MOD,MOD-); b=(b+MOD)%MOD;
     LL ans=;
     for(size_t i = ; i < n; ++i) if (s[i]==''||s[i]=='') ans=(ans+quick(,i))%MOD;
     cout<<a*b%MOD*ans%MOD<<endl;
     return EXIT_SUCCESS;
 }                /* ----------  end of function main  ---------- */

搞定，收工