hdu 6070 Dirt Ratio二分 线段树

Dirt Ratio

Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)

Problem Description

In ACM/ICPC contest, the ''Dirt Ratio'' of a team is calculated in the following way. First let's ignore all the problems the team didn't pass, assume the team passed Xproblems during the contest, and submitted Y times for these problems, then the ''Dirt Ratio'' is measured as XY. If the ''Dirt Ratio'' of a team is too low, the team tends to cause more penalty, which is not a good performance.



Picture from MyICPC


Little Q is a coach, he is now staring at the submission list of a team. You can assume all the problems occurred in the list was solved by the team during the contest. Little Q calculated the team's low ''Dirt Ratio'', felt very angry. He wants to have a talk with them. To make the problem more serious, he wants to choose a continuous subsequence of the list, and then calculate the ''Dirt Ratio'' just based on that subsequence.

Please write a program to find such subsequence having the lowest ''Dirt Ratio''.

 

Input

The first line of the input contains an integer T(1≤T≤15), denoting the number of test cases.

In each test case, there is an integer n(1≤n≤60000) in the first line, denoting the length of the submission list.

In the next line, there are n positive integers a1,a2,...,an(1≤ai≤n), denoting the problem ID of each submission.

 

Output

For each test case, print a single line containing a floating number, denoting the lowest ''Dirt Ratio''. The answer must be printed with an absolute error not greater than 10−4.

 

Sample Input

151 2 1 2 3

 

Sample Output

0.5000000000
Hint
 For every problem, you can assume its final submission is accepted.
 

 

Source

2017 Multi-University Training Contest - Team 4

对于区间[L ,R] ，不同数有X个， f = X/(R-L+1)  ， 求最小的f 。

首先二分f 

对于给定的f ,

枚举右端点r 。 有 X / (r-l+1) <= f 

X + l * f <= (r+1) * f 

这里对于给定的r , (r+1)*f 为定值。

转换为寻找最小的X+l*f 

对于不同的左端点，其实l*f 也是定值。

转换为维护X的值，即：维护[1,r] ,[2,r],[3,r]....[r,r]区间不同数的个数，可用线段树来维护。

import java.io.BufferedReader;import java.io.InputStream;import java.io.InputStreamReader;import java.io.PrintWriter;import java.math.BigInteger;import java.util.Arrays;import java.util.StringTokenizer;public class Main {public static void main(String[] args) {new Task().solve();}}class Task {InputReader in = new InputReader(System.in) ;PrintWriter out = new PrintWriter(System.out) ;int n ; int[] num ;double[] min , add ;int[] last = new int[60008]  ;void build(int l , int r , int t , double mid){add[t] = 0 ;if(l == r){min[t] = l * mid ;return ;}int m = (l + r) >> 1 ;build(l, m, t<<1 , mid) ;build(m+1, r, t<<1|1 , mid) ;up(t) ;}void down(int t){if(add[t] != 0){add[t<<1] += add[t] ;add[t<<1|1] += add[t] ;min[t<<1] += add[t] ;min[t<<1|1] += add[t] ;add[t] = 0 ;}}void up(int t){min[t] = Math.min(min[t<<1] , min[t<<1|1]) ;}void update(int L , int R , double d , int l , int r , int t){if(L <= l && r <= R){min[t] += d ;add[t] += d ; return ;}        down(t) ;int m = (l + r) >> 1 ;if(L <= m){update(L, R, d, l, m, t<<1) ;}if(R > m){update(L, R, d, m+1, r, t<<1|1) ;}up(t) ;}double query(int L , int R , int l , int r , int t){if(L <= l && r <= R){return min[t] ;}down(t) ;int m = (l + r) >> 1 ;double reslut = Double.MAX_VALUE ;if(L <= m){reslut = Math.min(reslut, query(L, R, l, m, t<<1)) ;}if(R > m){reslut = Math.min(reslut, query(L, R, m+1, r, t<<1|1)) ;}up(t) ;return reslut ;}boolean judge(double mid){Arrays.fill(last, 0) ;build(1, n, 1 , mid) ;for(int r = 1 ; r <= n ; r++){update(last[num[r]]+1, r, 1 , 1 , n , 1) ;    double q = query(1, r, 1, n, 1) ;    if(q <= mid * (r+1)){    return true ;    }    last[num[r]] = r ;}return false ; }void solve(){int t = in.nextInt() ;while(t-- > 0){ n = in.nextInt() ; num = new int[n+1] ; min = new double[n<<2] ; add = new double[n<<2] ; for(int i = 1 ; i <= n ; i++){ num[i] = in.nextInt() ; } double l = 0.0 , r = 1.0 , result = -1 ; for(int step = 0 ; step < 20 ; step++){ double mid = (l + r) * 0.5 ; if(judge(mid)){ r = mid ; result = mid ;  } else{ l = mid ; } } out.printf("%.10f%n" , result) ;} out.flush() ;}}class InputReader {        public BufferedReader reader;        public StringTokenizer tokenizer;            public InputReader(InputStream stream) {            reader = new BufferedReader(new InputStreamReader(stream), 32768);            tokenizer = new StringTokenizer("");        }            private void eat(String s) {            tokenizer = new StringTokenizer(s);        }            public String nextLine() {             try {                return reader.readLine();            } catch (Exception e) {                return null;            }        }            public boolean hasNext() {            while (!tokenizer.hasMoreTokens()) {                String s = nextLine();                if (s == null)                    return false;                eat(s);            }            return true;        }            public String next() {            hasNext();            return tokenizer.nextToken();        }            public int nextInt() {            return Integer.parseInt(next());        }            public int[] nextInts(int n) {            int[] nums = new int[n];            for (int i = 0; i < n; i++) {                nums[i] = nextInt();            }            return nums;        }            public long nextLong() {            return Long.parseLong(next());        }            public double nextDouble() {            return Double.parseDouble(next());        }            public BigInteger nextBigInteger() {            return new BigInteger(next());        }        }    

转换