B - Dice
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
Submit Status Practice HDU 4652
Description
You have a dice with m faces, each face contains a distinct number. We assume when we tossing the dice, each face will occur randomly and uniformly. Now you have T query to answer, each query has one of the following form:
0 m n: ask for the expected number of tosses until the last n times results are all same.
1 m n: ask for the expected number of tosses until the last n consecutive results are pairwise different.
Input
The first line contains a number T.(1≤T≤100) The next T line each line contains a query as we mentioned above. (1≤m,n≤10 6) For second kind query, we guarantee n≤m. And in order to avoid potential precision issue, we guarantee the result for our query will not exceeding 109 in this problem.
Output
For each query, output the corresponding result. The answer will be considered correct if the absolute or relative error doesn‘t exceed 10-6.
Sample Input
6
0 6 1
0 6 3
0 6 5
1 6 2
1 6 4
1 6 6
10
1 4534 25
1 1232 24
1 3213 15
1 4343 24
1 4343 9
1 65467 123
1 43434 100
1 34344 9
1 10001 15
1 1000000 2000
Sample Output
1.000000000
43.000000000
1555.000000000
2.200000000
7.600000000
83.200000000
25.586315824
26.015990037
15.176341160
24.541045769
9.027721917
127.908330426
103.975455253
9.003495515
15.056204472
4731.706620396
先附个大牛的网址:http://blog.csdn.net/ten_three/article/details/18924459
第一个图片中最后四，五行中的 1/m 应为 m .... 图片里的不对
#include <cstdio>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
double solve0(int m,int n)
{
double ans=0;
for(int i=0;i<=n-1;i++)
ans+=pow(1.0*m,i);
return ans;
}
double solve1(int m,int n)
{
double ans=0;
double tmp=1;
for(int i=1;i<=n;i++)
{
ans+=tmp;
tmp*=(m+0.0)/(m-i);
}
return ans;
}
int main()
{
int t;
while(scanf("%d",&t)!=EOF)
{
while(t--)
{
int n,m,op;
scanf("%d%d%d",&op,&m,&n);
if(op==0)
printf("%.9lf\n",solve0(m,n));
else
printf("%.9lf\n",solve1(m,n));
}
}
return 0;
}