ZS and The Birthday Paradox

Time Limit: 20 Sec  Memory Limit: 512 MB

Description

  

Input

  

Output

  

Sample Input

  4 3

Sample Output

  23 128

HINT

  

Solution

  

Code

 #include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<vector>
using namespace std;
typedef long long s64; const int ONE = ;
const int MOD = 1e6 + ; s64 n, k;
s64 num = ;
s64 Fz = , Fm = ; int get()
{
int res;char c;
while( (c=getchar())< || c> );
res=c-;
while( (c=getchar())>= && c<= )
res=res*+c-;
return res;
} s64 Quickpow(s64 a, s64 b)
{
s64 res = ;
while(b)
{
if(b & ) res = res * a % MOD;
a = a * a % MOD;
b >>= ;
}
return res;
} int main()
{
cin>>n>>k; num = ;
while((1LL << num) < k) num++;
if(num > n) {printf("1 1"); return ;} num = ;
for(s64 i = k - ; i >= ; i >>= )
num += (i >> ); s64 a = Quickpow(, n); Fz = ;
for(int i = ; i < k; i++)
{
Fz = (s64)Fz * (a - i + MOD) % MOD;
if(Fz == ) break;
}
Fm = Quickpow(a, k - ); int inv = Quickpow(Quickpow(, num), MOD - ); Fz = (s64)Fz * inv % MOD, Fm = (s64)Fm * inv % MOD;
Fz = (Fm - Fz + MOD) % MOD; cout<<Fz<<" "<<Fm<<endl;
}

【Codeforces711E】ZS and The Birthday Paradox [数论]的更多相关文章

  1. CF369E. ZS and The Birthday Paradox

    /* cf369E. ZS and The Birthday Paradox http://codeforces.com/contest/711/problem/E 抽屉原理+快速幂+逆元+勒让德定理 ...

  2. codeforces 711E E. ZS and The Birthday Paradox(数学+概率)

    题目链接: E. ZS and The Birthday Paradox. time limit per test 2 seconds memory limit per test 256 megaby ...

  3. ZS and The Birthday Paradox

    ZS and The Birthday Paradox 题目链接:http://codeforces.com/contest/711/problem/E 数学题(Legendre's formula) ...

  4. Codeforces 711E ZS and The Birthday Paradox 数学

    ZS and The Birthday Paradox 感觉里面有好多技巧.. #include<bits/stdc++.h> #define LL long long #define f ...

  5. Codeforces Round #369 (Div. 2) E. ZS and The Birthday Paradox 数学

    E. ZS and The Birthday Paradox 题目连接: http://www.codeforces.com/contest/711/problem/E Description ZS ...

  6. Codeforces 711E ZS and The Birthday Paradox

    传送门 time limit per test 2 seconds memory limit per test 256 megabytes input standard input output st ...

  7. cf711E ZS and The Birthday Paradox

    ZS the Coder has recently found an interesting concept called the Birthday Paradox. It states that g ...

  8. 【28.57%】【codeforces 711E】ZS and The Birthday Paradox

    time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...

  9. codeforces 711E. ZS and The Birthday Paradox 概率

    已知一年365天找23个人有2个人在同一天生日的概率 > 50% 给出n,k ,表示现在一年有2^n天,找k个人,有2个人在同一天生日的概率,求出来的概率是a/b形式,化到最简形式,由于a,b可 ...

随机推荐

  1. FromHandle临时对象一探究竟

    我们在调用CWnd::GetDlgItem()函数时,MSDN告诉我们:The returned pointer may be temporary and should not be stored f ...

  2. new关键字 、this关键字、base关键字

    使用new,所做的三件事: 1. (类是引用对象,引用对象是在堆中开辟空间)在堆中开辟空间 2. 在开辟的堆空间中创建对象 3. 调用对象的构建函数 4. 隐藏父类成员:子类的成员可以与隐藏从父类继承 ...

  3. jenkins部署springboot多项目

    war包的部署问题不大,这里记录jar包的部署过程: 1:jar包的体积过大问题 pom.xml参考以下配置(依赖包会分离到target/lib/,jar包体积由几十M缩小到几k) <build ...

  4. [计算机网络-应用层] HTTP协议

    1.HTTP概况 Web的应用层协议是超文本传输协议(HTTP),它是Web的核心. HTTP由两部分程序实现:一个客户机程序和一个服务器程序,它们运行在不同的端系统中,通过交换HTTP报文进行对话. ...

  5. asp.netMVC中实现分页方法

    方法一:使用传统的sql语句实现分页,    public class UserprintDao如下 /// <summary> /// 取得用户申请记录列表(按分页) /// </ ...

  6. 最小角回归 LARS算法包的用法以及模型参数的选择(R语言 )

    Lasso回归模型,是常用线性回归的模型,当模型维度较高时,Lasso算法通过求解稀疏解对模型进行变量选择.Lars算法则提供了一种快速求解该模型的方法.Lars算法的基本原理有许多其他文章可以参考, ...

  7. BZOJ4552:[HEOI2016/TJOI2016]排序——题解

    https://www.lydsy.com/JudgeOnline/problem.php?id=4552 https://www.luogu.org/problemnew/show/P2824 在2 ...

  8. BZOJ3223:文艺平衡树——超详细题解

    http://www.lydsy.com/JudgeOnline/problem.php?id=3223 题面复制于洛谷. 题目背景 这是一道经典的Splay模板题——文艺平衡树. 题目描述 您需要写 ...

  9. Educational Codeforces Round 39 (Rated for Div. 2) G

    Educational Codeforces Round 39 (Rated for Div. 2) G 题意: 给一个序列\(a_i(1 <= a_i <= 10^{9}),2 < ...

  10. 洛谷 P2657 [SCOI2009]windy数 解题报告

    P2657 [SCOI2009]windy数 题目描述 \(\tt{windy}\)定义了一种\(\tt{windy}\)数.不含前导零且相邻两个数字之差至少为\(2\)的正整数被称为\(\tt{wi ...