FZU - 2037 -Maximum Value Problem(规律题)

最新推荐文章于 2021-05-05 21:35:05 发布

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最新推荐文章于 2021-05-05 21:35:05 发布

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原文链接：http://www.cnblogs.com/Staceyacm/p/10840355.html

本文探讨了一个经典的数组中寻找最大值的问题，通过分析不同排列下“max=v”语句的执行次数，提出了一种算法来计算所有可能排列中该语句的总执行次数及期望执行次数。输入为数组长度，输出包括总执行次数模1,000,000,007和期望执行次数。

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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

2
2
3

Sample Output

Case 1: 3 1.500000
Case 2: 11 1.833333

思路；第n项的交换次数为F[n]=(n-1)!+F[n-1]*n;后面的为res[n]=1.0/n+res[n-1];
预处理一下输出就行了
代码：

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<vector>
#include<cmath>

const int maxn=1e5+5;
const int mod=1e9+7;
typedef long long ll;
using namespace std;
ll f[10*maxn];
double res[10*maxn];
int main()
{
    ll a=1;
    f[0]=0;
    f[1]=1;
    res[1]=1;
    for(int t=2;t<=1000000;t++)
    {
        f[t]=((a*(t-1))%mod+((t)*f[t-1])%mod)%mod;
        a=(a*(t-1))%mod;
        res[t]=1.0/t+res[t-1];
        //printf("%.6f\n",res[t]);
    }
    int T;
    int  n;
    cin>>T;
    int cnt=1;
    while(T--)
    {
      scanf("%d",&n);
      
      printf("Case %d: %d ",cnt++,f[n]);
      printf("%.6f\n",res[n]);
      
    }
    return 0;
}

转载于:https://www.cnblogs.com/Staceyacm/p/10840355.html