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可能非常大,对a,b分别%(10^6+3) 注意,这道题是先化到最简,再分别取模 首先,特判 k > 2^n 时,a = 1,b = 1 没有2个人同一天生日的概率为: 2^n * (2^n - 1) * ... * (2^n - k + 1) / 2^(nk) 所以a,b化简之前分别是: a = 2nk…

ZS and The Birthday Paradox 感觉里面有好多技巧.. #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ(x…

传送门 time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output ZS the Coder has recently found an interesting concept called the Birthday Paradox. It states that given a random set of \(23\) people,…

[题目链接] http://codeforces.com/problemset/problem/711/E [题目大意] 假设一年有2^n天,问k个小朋友中有两个小朋友生日相同的概率. 假设该概率约分后为 p / q ,输出p , q对1000003取模的解. [题解] 当k比天数要大时是肯定成立的,否则答案为1-A(2n,k) / (2n)k, 考虑A(2n,k)=2n*(2n-1)*……*(2n-k+1),所以分子和分母的最大公约数是2的幂次,暴力计算分子分母,以及计算最大公约数的逆元,就可…

题目链接: E. ZS and The Birthday Paradox. time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output ZS the Coder has recently found an interesting concept called the Birthday Paradox. It states that giv…

E. ZS and The Birthday Paradox 题目连接: http://www.codeforces.com/contest/711/problem/E Description ZS the Coder has recently found an interesting concept called the Birthday Paradox. It states that given a random set of 23 people, there is around 50% c…

ZS and The Birthday Paradox 题目链接:http://codeforces.com/contest/711/problem/E 数学题(Legendre's formula) 这题是以生日悖论(如果有23个或23个以上的人,那么至少有两个人的生日相同的概率要大于50%)为背景的数学题.公式本身不难推,主要是对大数的处理. 首先,我们需要找到分子和分母的公因数,这里用到了Legendre's formula(举个例子:10!的因数中质数3的个数为[10/3+10/(3^2…

/* 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+费马小定理+欧拉定理+数论 题解:https://amoshyc.github.io/ojsolution-build/cf/cf369/pe.html 坑点: 1.long long 类型的常量一定要加LL,否则1<<n只在int范围内 2.带模的题目,最后一定要判断是否答案为负,答案为负数要…

除非特别忙,我接下来会尽可能翻译我做的每道CF题的题面! Codeforces 148D 一袋老鼠 Bag of mice | 概率DP 水题 题面 胡小兔和司公子都认为对方是垃圾. 为了决出谁才是垃圾,大哥拿来了一袋老鼠,其中有w只白老鼠和b只黑老鼠.胡小兔先抓,先抓到白老鼠的人赢. 每次学姐抓完老鼠之后,总会有另外一只老鼠从袋子里自己跑出来(这只老鼠不算任何人抓的),而胡小兔抓老鼠时则不会发生这样的事. 每次袋子里的每只老鼠被抓到的概率相等,当有一只老鼠跑出来的时候,每只老鼠跑出来的几率也相…

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> #includ…