Card
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 191 Accepted Submission(s): 52
Special Judge
Problem Description
There are x cards on the desk, they are numbered from 1 to x. The score of the card which is numbered i(1<=i<=x) is i. Every round BieBie picks one card out of the x cards,then puts it back. He does the same operation for b rounds. Assume that the score of the j-th card he picks is Sj . You are expected to calculate the expectation of the sum of the different score he picks.
Input
Multi test cases,the first line of the input is a number T which indicates the number of test cases.
In the next T lines, every line contain x,b separated by exactly one space.
[Technique specification]
All numbers are integers.
1<=T<=500000
1<=x<=100000
1<=b<=5
In the next T lines, every line contain x,b separated by exactly one space.
[Technique specification]
All numbers are integers.
1<=T<=500000
1<=x<=100000
1<=b<=5
Output
Each case occupies one line. The output format is Case #id: ans, here id is the data number which starts from 1,ans is the expectation, accurate to 3 decimal places.
See the sample for more details.
See the sample for more details.
Sample Input
2
2 3
3 3
Sample Output
Case #1: 2.625
Case #2: 4.222
View Code
Hint
For the first case, all possible combinations BieBie can pick are (1, 1, 1),(1,1,2),(1,2,1),(1,2,2),(2,1,1),(2,1,2),(2,2,1),(2,2,2)
For (1,1,1),there is only one kind number i.e. 1, so the sum of different score is 1.
However, for (1,2,1), there are two kind numbers i.e. 1 and 2, so the sum of different score is 1+2=3.
So the sums of different score to corresponding combination are 1,3,3,3,3,3,3,2
So the expectation is (1+3+3+3+3+3+3+2)/8=2.625我的做法是把它想象成一棵 m 层的 x 叉树(底层有 x^b 个叶子结点), 然后计算每个数字(1~x )要加的次数。
对于( i = 1 ~ x ) ..
在第1层就要加 x^( m-1 ) 次 。
第2层就要加 x^(m-2) *(x-1) 次 。
....
第 i 层就要加 x^( m - i ) * (x-1)^( i - 1 )。
....
第m层就要加 x^(0) *(x-1)^(m-1) 次。
以 i = 1 为例 ,如下图:
那么每个数加的次数就是 x^i * ( x - 1 )^( m - i - 1 ) 次 [ 0 <= i < m ]。
总共加的和 sum = sigma(j) * x^i * ( x - 1 )^( m - i - 1 ) 次 [ 1<=j <= x , 0 <= i < m ]。
sigma(j) [1<=j <= x ] = (1+x)*x/2。
那么 sum = (1+x) * x / 2 * x^i * ( x - 1 )^( m - i - 1 ) , [ 0 <= i < m ] 。
那么期望 Ex = sum / ( x ^ n ) 。
官方题解是给出普通用概率方法求 :
设Xi代表分数为i的牌在b次操作中是否被选到,Xi=1为选到,Xi=0为未选到 那么期望EX=1*X1+2*X2+3*X3+…+x*Xx Xi在b次中被选到的概率是1-(1-1/x)^b 那么E(Xi)= 1-(1-1/x)^b 那么EX=1*E(X1)+2*E(X2)+3*E(X3)+…+x*E(Xx)=(1+x)*x/2*(1-(1-1/x)^b)
#include <iostream> #include <cstdio> #include <cstring> #include <string> #include <cmath> #include <vector> #include <queue> #include <map> #include <set> #include <stack> #include <algorithm> using namespace std; #define root 1,n,1 #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define lr rt<<1 #define rr rt<<1|1 typedef long long LL; typedef pair<int,int>pii; #define X first #define Y second const int oo = 1e9+7; const double PI = acos(-1.0); const double eps = 1e-6 ; const int N = 100010; double n ;int m ; void Run() { scanf("%lf%d",&n,&m); double avg = ( 1.0 + n ) * n / 2.0 * pow( 1.0 / n , (double) m ) , res = 0 ; for( int i = 0 ; i < m ; ++i ) { res += avg * pow( n - 1.0 , (double)i )*pow( (double)n, m-i-1.0 ); } printf("%.3lf ",res); } int main() { #ifdef LOCAL freopen("in.txt","r",stdin); #endif // LOCAL ios::sync_with_stdio(false); int _ , cas = 1 ; scanf("%d",&_); while( _-- ){ printf("Case #%d: ",cas++); Run(); } }