poj 6243 Dogs and Cages

Dogs and Cages

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 56    Accepted Submission(s): 40
Special Judge

Problem Description

Jerry likes dogs. He has N

dogs numbered 0,1,...,N−1

. He also has N

cages numbered 0,1,...,N−1

. Everyday he takes all his dogs out and walks them outside. When he is back
home, as dogs can’t recognize the numbers, each dog just randomly selects a cage
and enters it. Each cage can hold only one dog.
One day, Jerry noticed that
some dogs were in the cage with the same number of themselves while others were
not. Jerry would like to know what’s the expected number of dogs that are NOT in
the cage with the same number of themselves.

 

Input

The first line of the input gives the number of test
cases, T

. T

test cases follow.
Each test case contains only one number N

, indicating the number of dogs and cages.
1≤T≤105

1≤N≤105

 

Output

For each test case, output one line containing “Case
#x: y”, where x

is the test case number (starting from 1) and y

is the expected number of dogs that are NOT in the cage with the same number of
itself.
y

will be considered correct if it is within an absolute or relative error of
10−6

of the correct answer.

 

Sample Input

2
1
2

 

Sample Output

Case #1: 0.0000000000
Case #2: 1.0000000000

Hint

In the first test case, the only dog will enter the only cage. So the answer is 0.
In the second test case, if the first dog enters the cage of the same number, both dogs are in the cage of the same number,
the number of mismatch is 0. If both dogs are not in the cage with the same number of itself, the number of mismatch is 2.
So the expected number is (0+2)/2=1.

 

题意：随机产生的长度为 n 的排列 {Ai}，求期望有多少位置满足 A[i] != i。

思路：和的期望等于期望的和；可以考虑每个位置对总期望的贡献，每个位置满足A[i]!=i有n-1种情况，满足条件的期望为(n-1)/n;

一共n个位置，故总期望为n-1;

AC代码：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<cstdio>
#include<vector>
#include<algorithm>
#include<cstring>
#include<string>
#include<queue>
#include<cmath>
using namespace std;
#define INF 0x3f3f3f3f
typedef vector<double> vec;
typedef vector<vec> mat;
const int N_MAX = ;
double n;
int T;

int main() {
    scanf("%d", &T);
    int k = ;
    while (T--) {
        k++;
        scanf("%lf",&n);
        printf("Case #%d: %.10f\n",k,n-);
    }
    return ;
}