FZU 2037

 

Let’s start with a very classical problem. Given an array a[1…n] of positive numbers, if the value of each element in the array is distinct, how to find the maximum element in this array? You may write down the following pseudo code to solve this problem:

function find_max(a[1…n])

max=0;

for each v from a

if(max<v)

max=v;

return max;

However, our problem would not be so easy. As we know, the sentence ‘max=v’ would be executed when and only when a larger element is found while we traverse the array. You may easily count the number of execution of the sentence ‘max=v’ for a given array a[1…n].

Now, this is your task. For all permutations of a[1…n], including a[1…n] itself, please calculate the total number of the execution of the sentence ‘max=v’. For example, for the array [1, 2, 3], all its permutations are [1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2] and [3, 2, 1]. For the six permutations, the sentence ‘max=v’ needs to be executed 3, 2, 2, 2, 1 and 1 times respectively. So the total number would be 3+2+2+2+1+1=11 times.

Also, you may need to compute that how many times the sentence ‘max=v’ are expected to be executed when an array a[1…n] is given (Note that all the elements in the array is positive and distinct). When n equals to 3, the number should be 11/6= 1.833333.

Input

The first line of the input contains an integer T(T≤100,000), indicating the number of test cases. In each line of the following T lines, there is a single integer n(n≤1,000,000) representing the length of the array.

Output

For each test case, print a line containing the test case number (beginning with 1), the total number mod 1,000,000,007

and the expected number with 6 digits of precision, round half up in a single line.

Sample Input

223

Sample Output

Case 1: 3 1.500000Case 2: 11 1.833333

 第一个答案公式f(n) = (n-1)!+f(n-1)+(n-1)*f(n-1);    

 第二个答案公式 f(n) = f(n-1)+1/n; （由第一个 公式 除于 n！）

1.如果n放在最后，原来的f(n-1)依旧存在不受影响，同时加多（n-1)!个
2.接着对于（n-1)！个排列，把n这个数分别放在1,2,3，4，n-1这些位置，如下f(3)->f(4)

3个数的排列：

123

132

213

231

312

4个数的排列可以看成在3个数排列的基础上添加一位，则4可以放在第一，第二，第三，第四位

①如果放在最后一位则有 (4-1)! 个

1234

1324

2134

2314

3124

3214

②如果是放在第一，第二和第三时则有 （4-1）*f(4-1);

4123   1423   1243

4132   1432   1342

4213   2413   2143

4231   2431   2341

4312   3412   3142

4321   3421   3241 

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#define MAX 1000005#define mod 1000000007#define  ll long long intusing namespace std;ll a[MAX], b[MAX];double c[MAX];int n, t;int main() {scanf("%d",&n);a[1] = 1,b[1] = 1,c[1] = 1;for (int i = 2; i < MAX; i++) {a[i] = (a[i-1] *i) % mod;b[i] = (a[i-1] + i*b[i-1]) % mod;c[i] = c[i-1] + 1.0 / i;}for (int i = 1; i <=n; i++)  {scanf("%d",&t);printf("Case %d: %lld %.6lf\n",i,b[t],c[t]);}return 0;}