Codeforces 621C Wet Shark and Flowers 【期望】
来源:互联网 发布:mac子弹头 dubonnet 编辑:程序博客网 时间:2024/06/05 04:17
There are n sharks who grow flowers for Wet Shark. They are all sitting around the table, such that sharks i and i + 1 are neighbours for alli from 1 to n - 1. Sharks n and 1 are neighbours too.
Each shark will grow some number of flowers si. For i-th shark value si is random integer equiprobably chosen in range from li to ri. Wet Shark has it's favourite prime number p, and he really likes it! If for any pair of neighbouring sharks i and j the product si·sj is divisible by p, then Wet Shark becomes happy and gives 1000 dollars to each of these sharks.
At the end of the day sharks sum all the money Wet Shark granted to them. Find the expectation of this value.
The first line of the input contains two space-separated integers n and p (3 ≤ n ≤ 100 000, 2 ≤ p ≤ 109) — the number of sharks and Wet Shark's favourite prime number. It is guaranteed that p is prime.
The i-th of the following n lines contains information about i-th shark — two space-separated integers li and ri (1 ≤ li ≤ ri ≤ 109), the range of flowers shark i can produce. Remember that si is chosen equiprobably among all integers from li to ri, inclusive.
Print a single real number — the expected number of dollars that the sharks receive in total. You answer will be considered correct if its absolute or relative error does not exceed 10 - 6.
Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if .
3 21 2420 421420420 420421
4500.0
3 51 42 311 14
0.0
A prime number is a positive integer number that is divisible only by 1 and itself. 1 is not considered to be prime.
Consider the first sample. First shark grows some number of flowers from 1 to 2, second sharks grows from 420 to 421 flowers and third from 420420 to 420421. There are eight cases for the quantities of flowers (s0, s1, s2) each shark grows:
- (1, 420, 420420): note that s0·s1 = 420, s1·s2 = 176576400, and s2·s0 = 420420. For each pair, 1000 dollars will be awarded to each shark. Therefore, each shark will be awarded 2000 dollars, for a total of 6000 dollars.
- (1, 420, 420421): now, the product s2·s0 is not divisible by 2. Therefore, sharks s0 and s2 will receive 1000 dollars, while shark s1 will receive 2000. The total is 4000.
- (1, 421, 420420): total is 4000
- (1, 421, 420421): total is 0.
- (2, 420, 420420): total is 6000.
- (2, 420, 420421): total is 6000.
- (2, 421, 420420): total is 6000.
- (2, 421, 420421): total is 4000.
The expected value is .
In the second sample, no combination of quantities will garner the sharks any money.
题意:有n个人围成一圈,已知每个人都有一个选数区间[l, r]可以等概率的在区间里面选数。若x 和 (x+1) % n选数之和% p == 0,那么每人将会获得1000元。问获得总钱数的期望。
思路:定义P[i]为第i和(i+1) % n获得金钱的概率,期望E[i]为P[i] * 2000。期望的线性性质,总期望 = sigma(E[i]) (1 <= i <= n)。
AC代码:坑啊,还以为保留一位小数。。。
#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <set>#include <vector>#include <string>#define INF 1000000#define eps 1e-8#define MAXN (200000+10)#define MAXM (100000+10)#define Ri(a) scanf("%d", &a)#define Rl(a) scanf("%lld", &a)#define Rf(a) scanf("%lf", &a)#define Rs(a) scanf("%s", a)#define Pi(a) printf("%d\n", (a))#define Pf(a) printf("%.2lf\n", (a))#define Pl(a) printf("%lld\n", (a))#define Ps(a) printf("%s\n", (a))#define W(a) while((a)--)#define CLR(a, b) memset(a, (b), sizeof(a))#define MOD 1000000007#define LL long long#define lson o<<1, l, mid#define rson o<<1|1, mid+1, r#define ll o<<1#define rr o<<1|1#define PI acos(-1.0)#pragma comment(linker, "/STACK:102400000,102400000")#define fi first#define se secondusing namespace std;typedef pair<int, int> pii;int main(){ int n, p; Ri(n); Ri(p); int total1, total2, num1, num2; int l, r; Ri(l); Ri(r); int total = total1 = r - l + 1; int num = num1 = (total - (r / p - (l-1) / p)); double ans = 0; for(int i = 2; i <= n; i++) { Ri(l); Ri(r); total2 = r - l + 1; num2 = total2 - (r / p - (l-1) / p); ans += 1.0 * (1 - num1*1.0 / total1 * num2*1.0 / total2) * 2000; num1 = num2; total1 = total2; } ans += 1.0 * (1 - num1*1.0 / total1 * num*1.0 / total) * 2000; printf("%.10lf\n", ans); return 0;}
- Codeforces 621C Wet Shark and Flowers 【期望】
- 【概率与期望】[CodeForces - 621C]Wet Shark and Flowers
- CodeForces 621 C. Wet Shark and Flowers(期望)
- Codeforces 621C Wet Shark and Flowers
- CodeForces 621C Wet Shark and Flowers
- Codeforce 题目621C Wet Shark and Flowers(期望)
- Codeforces 621B Wet Shark and Flowers
- Codeforces Round #341 (Div. 2) C. Wet Shark and Flowers (期望)
- Codeforces Round #341 (Div. 2) C. Wet Shark and Flowers(期望)
- Codeforces Round #341 (Div. 2) C Wet Shark and Flowers - 期望与概率
- Codeforces Round #341 (Div. 2)C. Wet Shark and Flowers(数学期望)
- Codeforces Round #341 (Div. 2) C - Wet Shark and Flowers 数学期望
- 621C. Wet Shark and Flowers【概率】
- codeforce 621C Wet Shark and Flowers
- codeforces621C. Wet Shark and Flowers【求期望】
- Codeforces Round #341 (Div. 2) C. Wet Shark and Flowers
- Codeforces #341 C. Wet Shark and Flowers 概率论
- Codeforces Round #341 (Div. 2)--C. Wet Shark and Flowers
- 基本数据分析指标
- .Net在线付款---Paypal在线付款开发过程
- 使用xcode7(或是xcode6)遇到的一些问题
- mysql千万级数据库插入速度和读取速度的调整记录
- iOS手势操作简介(一)
- Codeforces 621C Wet Shark and Flowers 【期望】
- 系统分层架构设计参考
- 常用的一些验证
- iOS 定义具有位移操作的枚举的意义
- Linux下/proc目录简介
- 解压assets下的zip包并复制相应的文件
- Git
- 阿里开源Mysql分布式中间件:Cobar
- window下面搭建php集成环境xampp