HDU5159 Card

Card

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 429    Accepted Submission(s): 163
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

 

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.

 

Sample Input

22 33 3

 

Sample Output

Case #1: 2.625Case #2: 4.222
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 1，3，3，3，3，3，3，2So the expectation is (1+3+3+3+3+3+3+2)/8=2.625
  1.在每张牌中1~x中·每张牌出现的概率相同
  2. 1~x中 ，i在这b次中"有"出现的概率是(1-i从不出现) 即1-pow((x-1)/(1.0*x),b*1.0); 
#include<cstdio>#include<stack>#include<cstring>#include<algorithm>#include<iostream>#include<math.h>using namespace std;long long sum[100010];int main(){    //freopen("G://test.txt","r",stdin);    int m,n,t,x,b,count=0;    sum[0]=0;    for(int i=1;i<=100000;++i){    sum[i]=sum[i-1]+i;    }    scanf("%d",&t);    while(t--){        count++;    scanf("%d%d",&x,&b);    double fin;    double p=1-pow((x-1)/(1.0*x),b*1.0);      fin=sum[x]*p;    printf("Case #%d: %.3lf\n",count,fin);    }     return 0;}