zoukankan      html  css  js  c++  java
  • GYM100633J. Ceizenpok’s formula 扩展lucas模板

    J. Ceizenpok’s formula
    time limit per test
    2.0 s
    memory limit per test
    256 MB
    input
    standard input
    output
    standard output

    Dr. Ceizenp'ok from planet i1c5l became famous across the whole Universe thanks to his recent discovery — the Ceizenpok’s formula. This formula has only three arguments: n, k and m, and its value is a number of k-combinations of a set of n modulo m.

    While the whole Universe is trying to guess what the formula is useful for, we need to automate its calculation.

    Input

    Single line contains three integers n, k, m, separated with spaces (1 ≤ n ≤ 1018, 0 ≤ k ≤ n, 2 ≤ m ≤ 1 000 000).

    Output

    Write the formula value for given arguments n, k, m.

    Examples
    Input
    2 1 3
    Output
    2
    Input
    4 2 5
    Output
    1
     
     
     
    【题解】
    裸的扩展lucas + CRT,推荐博文http://blog.csdn.net/clove_unique/article/details/54571216
    因为特殊情况k = 0的时候,我以为是0,加了个特判,结果应该是1,不用特判。。。。。。。。。。卡了我半个晚上
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <cstdlib>
     5 #include <algorithm>
     6 #include <queue>
     7 #include <vector>
     8 #include <map>
     9 #include <string> 
    10 #include <cmath> 
    11 #include <sstream>
    12 #define min(a, b) ((a) < (b) ? (a) : (b))
    13 #define max(a, b) ((a) > (b) ? (a) : (b))
    14 #define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
    15 template<class T>
    16 inline void swap(T &a, T &b)
    17 {
    18     T tmp = a;a = b;b = tmp;
    19 }
    20 inline void read(long long &x)
    21 {
    22     x = 0;char ch = getchar(), c = ch;
    23     while(ch < '0' || ch > '9') c = ch, ch = getchar();
    24     while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
    25     if(c == '-') x = -x;
    26 }
    27 const long long INF = 0x3f3f3f3f;
    28 
    29 long long pow(long long a, long long b, long long mod)
    30 {
    31     long long r = 1, base = a % mod;
    32     for(;b;b >>= 1)
    33     {
    34         if(b & 1) r *= base, r %= mod;
    35         base *= base, base %= mod;
    36     }
    37     return r; 
    38 }
    39 void exgcd(long long a, long long b, long long &x, long long &y)
    40 {
    41     if(!b)x = 1, y = 0;
    42     else exgcd(b, a%b, y, x), y -= (a / b) * x;
    43 }
    44 long long ni(long long x, long long mod)
    45 {
    46     long long inv, y; exgcd(x, mod, inv, y);
    47     inv = (inv % mod + mod) % mod;
    48     if(!inv) inv = mod;
    49     return inv; 
    50 }
    51 int tiaoshi;
    52 long long calc(long long n, long long p, long long pt)
    53 {
    54     if(n == 0) return 1;
    55     long long ans = 1;
    56     for(long long i = 1;i <= pt;++ i) if(i % p) ans *= i, ans %= pt;
    57     ans = pow(ans, n/pt, pt);
    58     for(long long i = 1;i <= n%pt;++ i) 
    59         if(i % p) 
    60             ans *= i, ans %= pt;
    61     return ans * calc(n/p, p, pt) % pt;
    62 }
    63 long long C(long long n, long long m, long long p, long long pt)
    64 {
    65     if(n < m || n < 0 || m < 0) return 0;
    66     long long cnt = 0;
    67     for(long long i = n;i;i /= p) cnt += i/p;
    68     for(long long i = m;i;i /= p) cnt -= i/p;
    69     for(long long i = n - m;i;i /= p) cnt -= i/p;
    70     return pow(p, cnt, pt) * calc(n, p, pt) % pt * ni(calc(m, p, pt), pt) % pt * ni(calc(n - m, p, pt), pt) % pt;
    71 } 
    72 long long exlucas(long long n, long long m, long long mod)
    73 {
    74     long long tmp2 = mod, ans = 0;
    75     for(long long i = 2;i <= mod;++ i)
    76         if(tmp2 % i == 0)
    77         {
    78             long long pt = 1;
    79             while(tmp2 % i == 0)  tmp2 /= i, pt *= i;
    80             long long tmp3 = C(n, m, i, pt);
    81             ans += tmp3 * (mod/pt) % mod * ni(mod/pt, pt) % mod;
    82             ans %= mod;
    83         }
    84     return ans;
    85 }
    86 long long n,m,p;
    87 int main()
    88 {
    89     read(n), read(m), read(p);
    90     printf("%lld", exlucas(n, m, p)); 
    91     return 0;
    92 }
    GYM100633J
  • 相关阅读:
    vue2.0子组件修改父组件props数据的值
    Visual Studio串口通信与测控应用编程实践
    电梯安装与调试技术
    拾年
    人文生态视野下的城市景观形态研究
    生物真让人惊奇
    神奇生理科学美图大观
    藏在文物里的中国史2 夏商周
    思科UCS服务器统一计算
    Google改变生活
  • 原文地址:https://www.cnblogs.com/huibixiaoxing/p/8406941.html
Copyright © 2011-2022 走看看