else 最小 print math src 技術 date time integer

Dirt Ratio

Time Limit: 18000/9000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 2522 Accepted Submission(s): 1138
Special Judge

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 X

problems 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 1 5 1 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

求一個序列中的連續子序列S，使得 （S中不同元素的個數）/（S的長度）最小化。挺不錯的一道題，很考驗思維。對於這種最小化的題目很容易往二分上面想，得到 a/b<=k ,我們只要不斷二分k，找到下界即可，問題是對於k如何判定是否可行，枚舉所有區間顯然不靠譜，考慮這個式子 diff(l,r)/(r-l+1)<=k -> diff(l,r)+k*l<=k*(r+1), 我們可以枚舉一下右端點，然後找到一個最優的左端點判斷能否使得這個不等式成立即可，可以用線段樹區間修改來維護這個值，結點 u(L,R),保存的就是[L,R]中最小的 diff(l,r)+k*l ,顯然l€[L,R]。初始化根節點為k*L,然後遍歷右端點的時候更新對應的區間，也就是會對那些區間的diff造成變化。

　　

　　

  1 #include<iostream>
  2 #include<cstring>
  3 #include<queue>
  4 #include<cstdio>
  5 #include<stack>
  6 #include<set>
  7 #include<map>
  8 #include<cmath>
  9 #include<ctime>
 10 #include<time.h> 
 11 #include<algorithm>
 12 #include<bits/stdc++.h>
 13 using namespace std;
 14 #define mp make_pair
 15 #define pb push_back
 16 #define debug puts("debug")
 17 #define LL long long 
 18 #define pii pair<int,int>
 19 #define inf 0x3f3f3f3f
 20 
 21 #define mid ((L+R)>>1)
 22 #define lc (id<<1)
 23 #define rc (id<<1|1)
 24 #define eps 1e-5
 25 const int _M=60006;
 26 double sum[_M<<2];
 27 int laz[_M<<2];
 28 int a[_M],x[_M],pre[_M];
 29 void pushdown(int id){
 30     if(laz[id]){
 31         laz[lc]+=laz[id];
 32         laz[rc]+=laz[id];
 33         sum[lc]+=laz[id];
 34         sum[rc]+=laz[id];
 35         laz[id]=0;
 36     }
 37 }
 38 void pushup(int id){
 39     sum[id]=min(sum[lc],sum[rc]);
 40 }
 41 void build(int id,int L,int R,double k){
 42     if(L==R){
 43         sum[id]=k*L;
 44         return;
 45     }
 46     build(lc,L,mid,k);
 47     build(rc,mid+1,R,k);
 48     pushup(id);
 49 }
 50 void update(int id,int L,int R,int l,int r){
 51     if(L>=l&&R<=r){
 52         sum[id]++;
 53         laz[id]++;
 54         return ;
 55     }
 56     pushdown(id);
 57     if(l<=mid) update(lc,L,mid,l,r);
 58     if(r>mid) update(rc,mid+1,R,l,r);
 59     pushup(id);
 60 }
 61 double query(int id,int L,int R,int l,int r){
 62     if(L>=l&&R<=r){
 63         return sum[id];
 64     }
 65     pushdown(id);
 66     if(r<=mid) return query(lc,L,mid,l,r);
 67     else if(l>mid) return query(rc,mid+1,R,l,r);
 68     else return min(query(lc,L,mid,l,r),query(rc,mid+1,R,l,r));
 69     pushup(id);
 70 }
 71 bool ok(double k,int n){
 72     memset(sum,0,sizeof(sum));
 73     memset(laz,0,sizeof(laz));
 74     build(1,1,n,k);
 75     for(int i=1;i<=n;++i){
 76         update(1,1,n,pre[i]+1,i);
 77         if(query(1,1,n,1,i)<=k*(i+1)) return 1;
 78     }
 79     return 0;
 80 }
 81 int main(){
 82     int t,n,m,i,j;
 83     scanf("%d",&t);
 84     while(t--){
 85         scanf("%d",&n);
 86         memset(x,0,sizeof(x));
 87         for(i=1;i<=n;++i) {
 88             scanf("%d",a+i);
 89             pre[i]=x[a[i]];
 90             x[a[i]]=i;
 91         }
 92         
 93         double l=0,r=1.0;
 94         while(abs(l-r)>eps){
 95             double k=(l+r)/2;
 96             if(ok(k,n)) r=k;
 97             else l=k+eps;
 98         }
 99         printf("%.10f\n",l);
100     }
101     return 0;
102 }                                                                                       

HDU-6070-二分+線段樹