fzu 2037 Maximum Value Problem(规律？ 递推)

Problem Description

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

 Source

2011年全国大学生程序设计邀请赛（福州）

题意：给定n，求n全排列中，那max= 0从头去交换，交换到比他大为止。交换次数之和。

思路：没推出递推式。。打表找了规律推了公式f(n) = f(n - 1) * n + (n - 1)!。然后去递推。。注意求p[n]的时候p(n) = f(n) / (n)! = f(n -1)  / (n - 1)! + 1/n; 所以p(n) = p(n - 1) + 1/n;

代码：

#include <stdio.h>#include <string.h>#define MOD 1000000007const int N = 1000005;int t, n;long long f[N], jie[N];double p[N];void table() {    jie[0] = 1; f[0] = 0; p[0] = 0;    for (long long i = 1; i <= 1000000; i ++) {jie[i] = jie[i - 1] * i % MOD;f[i] = (f[i - 1] * i + jie[i - 1]) % MOD;p[i] = p[i - 1] + (1.0 / i);    }}int main() {    table();    scanf("%d", &t);    int cas = 0;    while (t--) {scanf("%d", &n);printf("Case %d: %lld %.6lf\n", ++cas, f[n], p[n]);    }    return 0;}