SPOJ FAVDICE

最新推荐文章于 2019-05-15 21:35:56 发布

原创最新推荐文章于 2019-05-15 21:35:56 发布 · 442 阅读

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Time Limit:391MS Memory Limit:1572864KB 64bit IO Format:%lld & %llu

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Description

BuggyD loves to carry his favorite die around. Perhaps you wonder why it's his favorite? Well, his die is magical and can be transformed into an N-sided unbiased die with the push of a button. Now BuggyD wants to learn more about his die, so he raises a question:

What is the expected number of throws of his die while it has N sides so that each number is rolled at least once?

Input

The first line of the input contains an integer t, the number of test cases. t test cases follow.

Each test case consists of a single line containing a single integer N (1 <= N <= 1000) - the number of sides on BuggyD's die.

Output

For each test case, print one line containing the expected number of times BuggyD needs to throw his N-sided die so that each number appears at least once. The expected number must be accurate to 2 decimal digits.

Example

Input:
2
1
12

Output:
1.00
37.24

概率题。

n/1+n/2+....+n/n;

#include <stdio.h>
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        double sum=0;
        for(int i=1;i<=n;i++)
            sum+=(double)n/i;
        printf("%.2lf\n",sum);
    }
    return 0;
}

另一种期望DP的做法。

#include <cstdio>
#include <cstring>
using namespace std;
#define N 1005
double dp[N];
int main()
{
        int T,k,n;
        scanf("%d",&T);
        while(T--)
        {
                scanf("%d%d",&k,&n);
                dp[1]=1.00;
                for(int i=2;i<=n;i++)
                        dp[i]=dp[i-1]+(k-dp[i-1])/k;
                printf("%.5f\n",dp[n]);
        }
     return 0;
}