ZOJ3557 How Many Sets II( Lucas定理)

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How Many Sets II

Time Limit: 2 Seconds      Memory Limit: 65536 KB

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 <= 104m <= 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

  1 /**
  2  * code generated by JHelper
  3  * More info: https://github.com/AlexeyDmitriev/JHelper
  4  * @author xyiyy @https://github.com/xyiyy
  5  */
  6 
  7 #include <iostream>
  8 #include <fstream>
  9 
 10 //#####################
 11 //Author:fraud
 12 //Blog: http://www.cnblogs.com/fraud/
 13 //#####################
 14 //#pragma comment(linker, "/STACK:102400000,102400000")
 15 #include <iostream>
 16 #include <sstream>
 17 #include <ios>
 18 #include <iomanip>
 19 #include <functional>
 20 #include <algorithm>
 21 #include <vector>
 22 #include <string>
 23 #include <list>
 24 #include <queue>
 25 #include <deque>
 26 #include <stack>
 27 #include <set>
 28 #include <map>
 29 #include <cstdio>
 30 #include <cstdlib>
 31 #include <cmath>
 32 #include <cstring>
 33 #include <climits>
 34 #include <cctype>
 35 
 36 using namespace std;
 37 #define rep(X, N) for(int X=0;X<N;X++)
 38 typedef long long ll;
 39 
 40 //
 41 // Created by xyiyy on 2015/8/15.
 42 //
 43 
 44 #ifndef ICPC_LUCAS_HPP
 45 #define ICPC_LUCAS_HPP
 46 
 47 //
 48 // Created by xyiyy on 2015/8/5.
 49 //
 50 
 51 #ifndef ICPC_INV_HPP
 52 #define ICPC_INV_HPP
 53 typedef long long ll;
 54 
 55 void extgcd(ll a, ll b, ll &d, ll &x, ll &y) {
 56     if (!b) {
 57         d = a;
 58         x = 1;
 59         y = 0;
 60     }
 61     else {
 62         extgcd(b, a % b, d, y, x);
 63         y -= x * (a / b);
 64     }
 65 }
 66 
 67 ll inv(ll a, ll mod) {
 68     ll x, y, d;
 69     extgcd(a, mod, d, x, y);
 70     return d == 1 ? (x % mod + mod) % mod : -1;
 71 }
 72 
 73 
 74 #endif //ICPC_INV_HPP
 75 
 76 
 77 ll C(int n, int m, ll mod) {
 78     if (n < m)return 0;
 79     if (m == 0)return 1;
 80     ll ret = 1;
 81     rep(i, m) {
 82         ret = ret * (n - i) % mod * inv(i + 1, mod) % mod;
 83     }
 84     return ret;
 85 }
 86 
 87 ll Lucas(ll n, ll m, ll mod) {
 88     if (m == 0)return 1;
 89     else return (C(n % mod, m % mod, mod) * Lucas(n / mod, m / mod, mod)) % mod;
 90 }
 91 
 92 
 93 #endif //ICPC_LUCAS_HPP
 94 
 95 class TaskI {
 96 public:
 97     void solve(std::istream &in, std::ostream &out) {
 98         int n, m, p;
 99         while (in >> n >> m >> p) {
100             out << Lucas(n - m + 1, m, p) % p << endl;
101         }
102     }
103 };
104 
105 int main() {
106     std::ios::sync_with_stdio(false);
107     std::cin.tie(0);
108     TaskI solver;
109     std::istream &in(std::cin);
110     std::ostream &out(std::cout);
111     solver.solve(in, out);
112     return 0;
113 }

 

posted on 2015-08-16 01:09  xyiyy  阅读(...)  评论(...编辑  收藏

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