ZOJ3557 How Many Sets II( Lucas定理)

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How Many Sets II

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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Given a set S = {1, 2, ..., n}, number m and p, your job is to count how many set T satisfies the following condition:

• T is a subset of S
• |T| = m
• T does not contain continuous numbers, that is to say x and x+1 can not both in T

Input

There are multiple cases, each contains 3 integers n ( 1 <= n <= 10⁹ ), m ( 0 <= m <= 10⁴, m <= n ) and p ( p is prime, 1 <= p <= 10⁹ ) in one line seperated by a single space, proceed to the end of file.

Output

Output the total number mod p.

Sample Input

5 1 11
5 2 11

Sample Output

5
6

这是个组合数的基础问题了，从n个数中挑出m个数，要求不相邻。显然公式就是C(n-m+1,m)，然后就可以直接用Lucas定理做了

 /**
  * code generated by JHelper
  * More info: https://github.com/AlexeyDmitriev/JHelper
  * @author xyiyy @https://github.com/xyiyy
  */

 #include <iostream>
 #include <fstream>

 //#####################
 //Author:fraud
 //Blog: http://www.cnblogs.com/fraud/
 //#####################
 //#pragma comment(linker, "/STACK:102400000,102400000")
 #include <iostream>
 #include <sstream>
 #include <ios>
 #include <iomanip>
 #include <functional>
 #include <algorithm>
 #include <vector>
 #include <string>
 #include <list>
 #include <queue>
 #include <deque>
 #include <stack>
 #include <set>
 #include <map>
 #include <cstdio>
 #include <cstdlib>
 #include <cmath>
 #include <cstring>
 #include <climits>
 #include <cctype>

 using namespace std;
 #define rep(X, N) for(int X=0;X<N;X++)
 typedef long long ll;

 //
 // Created by xyiyy on 2015/8/15.
 //

 #ifndef ICPC_LUCAS_HPP
 #define ICPC_LUCAS_HPP

 //
 // Created by xyiyy on 2015/8/5.
 //

 #ifndef ICPC_INV_HPP
 #define ICPC_INV_HPP
 typedef long long ll;

 void extgcd(ll a, ll b, ll &d, ll &x, ll &y) {
     if (!b) {
         d = a;
         x = ;
         y = ;
     }
     else {
         extgcd(b, a % b, d, y, x);
         y -= x * (a / b);
     }
 }

 ll inv(ll a, ll mod) {
     ll x, y, d;
     extgcd(a, mod, d, x, y);
     return d ==  ? (x % mod + mod) % mod : -;
 }

 #endif //ICPC_INV_HPP

 ll C(int n, int m, ll mod) {
     if (n < m)return ;
     if (m == )return ;
     ll ret = ;
     rep(i, m) {
         ret = ret * (n - i) % mod * inv(i + , mod) % mod;
     }
     return ret;
 }

 ll Lucas(ll n, ll m, ll mod) {
     if (m == )return ;
     else return (C(n % mod, m % mod, mod) * Lucas(n / mod, m / mod, mod)) % mod;
 }

 #endif //ICPC_LUCAS_HPP

 class TaskI {
 public:
     void solve(std::istream &in, std::ostream &out) {
         int n, m, p;
         while (in >> n >> m >> p) {
             out << Lucas(n - m + , m, p) % p << endl;
         }
     }
 };

 int main() {
     std::ios::sync_with_stdio(false);
     std::cin.tie();
     TaskI solver;
     std::istream &in(std::cin);
     std::ostream &out(std::cout);
     solver.solve(in, out);
     return ;
 }