Codeforces 621C Wet Shark and Flowers 【期望】
2016-02-01 14:04
232 查看
C. Wet Shark and Flowers
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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.
Input
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.
Output
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
.
Sample test(s)
input
output
input
output
Note
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。。。
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
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.
Input
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.
Output
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
.
Sample test(s)
input
3 2 1 2 420 421 420420 420421
output
4500.0
input
3 5 1 4 2 3 11 14
output
0.0
Note
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 second using 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; }
相关文章推荐
- jquery的淡入淡出效果
- 南京优步上线黄金区域,接单可享更高奖励!
- JAVA 发送下载文件
- alpha版、beta版、rc版的意思
- Sql日期时间格式转换
- Codeforces 621B Wet Shark and Bishops 【dp】
- Redis学习手册(Key操作命令)
- Codeforces 621A Wet Shark and Odd and Even 【水题】
- Mysql中count(*),DISTINCT的使用方法和效率研究
- E:Could not open lock file /var/lib/dpkg/lock - open (13:Permission denied) E:Unable to lock
- poj1258 Agri-Net(最小生成树)
- ASP.NET Core 1.0 Configuration 配置管理
- js中的逻辑与(&&)和逻辑或(||)
- 299. Bulls and Cows
- 启动Tomcat错误:he JRE_HOME environment variable is not
- swift-结构体
- Java返回距离当前时间段
- Take good care of your eyes
- Android-StrictMode-性能优化
- CSS中的绝对定位与相对定位