Codeforces Round #403 (Div. 2, based on Technocup 2017 Finals)

Andryusha is an orderly boy and likes to keep things in their place.

Today he faced a PRoblem to put his socks in the wardrobe. He has n distinct pairs of socks which are initially in a bag. The pairs are numbered from 1 to n. Andryusha wants to put paired socks together and put them in the wardrobe. He takes the socks one by one from the bag, and for each sock he looks whether the pair of this sock has been already took out of the bag, or not. If not (that means the pair of this sock is still in the bag), he puts the current socks on the table in front of him. Otherwise, he puts both socks from the pair to the wardrobe.

Andryusha remembers the order in which he took the socks from the bag. Can you tell him what is the maximum number of socks that were on the table at the same time?

The first line contains the single integer n (1 ≤ n ≤ 10⁵) — the number of sock pairs.

The second line contains 2n integers x₁, x₂, ..., x_2n (1 ≤ x_i ≤ n), which describe the order in which Andryusha took the socks from the bag. More precisely, x_i means that the i-th sock Andryusha took out was from pair x_i.

It is guaranteed that Andryusha took exactly two socks of each pair.

Print single integer — the maximum number of socks that were on the table at the same time.

11 1output
1input
32 1 1 3 2 3output
2Note
In the first example Andryusha took a sock from the first pair and put it on the table. Then he took the next sock which is from the first pair as well, so he immediately puts both socks to the wardrobe. Thus, at most one sock was on the table at the same time.
In the second example Andryusha behaved as follows:
Initially the table was empty, he took out a sock from pair 2 and put it on the table.Sock (2) was on the table. Andryusha took out a sock from pair 1 and put it on the table.Socks (1, 2) were on the table. Andryusha took out a sock from pair 1, and put this pair into the wardrobe.Sock (2) was on the table. Andryusha took out a sock from pair 3 and put it on the table.Socks (2, 3) were on the table. Andryusha took out a sock from pair 2, and put this pair into the wardrobe.Sock (3) was on the table. Andryusha took out a sock from pair 3 and put this pair into the wardrobe.Thus, at most two socks were on the table at the same time.
题意：先将每双鞋子的单只鞋子放到桌子上，凑够一双放进鞋柜，求桌子上鞋子最多的时候有几只鞋子。
#include<iostream>#include<cstdio>#include<algorithm>using namespace std;const int N = 100005;int a[N],b[N];int main(){    int n,x;    cin>>n;    int sum = 0,maxn = 1;    for(int i = 0;i < n*2;i++)    {        scanf("%d",&x);        b[x]++;        if(b[x] == 2)            sum--;        else            sum++;        maxn = max(maxn,sum);    }    printf("%d/n",maxn);    return 0;}
B. The Meeting Place Cannot Be Changedtime limit per test5 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output
The main road in Bytecity is a straight line from south to north. Conveniently, there are coordinates measured in meters from the southernmost building in north direction.
At some points on the road there are n friends, and i-th of them is standing at the point xi meters and can move with any speed no greater than vi meters per second in any of the two directions along the road: south or north.
You are to compute the minimum time needed to gather all the n friends at some point on the road. Note that the point they meet at doesn't need to have integer coordinate.
Input
The first line contains single integer n (2 ≤ n ≤ 60 000) — the number of friends.
The second line contains n integers x1, x2, ..., xn (1 ≤ xi ≤ 109) — the current coordinates of the friends, in meters.
The third line contains n integers v1, v2, ..., vn (1 ≤ vi ≤ 109) — the maximum speeds of the friends, in meters per second.
Output
Print the minimum time (in seconds) needed for all the n friends to meet at some point on the road.
Your answer will be considered correct, if its absolute or relative error isn't greater than 10 - 6. Formally, let your answer be a, while jury's answer be b. Your answer will be considered correct if  holds.
Examplesinput
37 1 31 2 1output
2.000000000000input
45 10 3 22 3 2 4output
1.400000000000Note
In the first sample, all friends can gather at the point 5 within 2 seconds. In order to achieve this, the first friend should go south all the time at his maximum speed, while the second and the third friends should go north at their maximum speeds.
题意：给你n个人的位置和从每个点出发的最大速度，让求最少的时间所有的人可以会合。
思路：二分，找接近的时间
#include<iostream>#include<cstdio>#include<algorithm>using namespace std;const double INF = 1e300;const int N = 6e4 + 5;int x[N],v[N];int main(){    int n;    cin >> n;    for(int i = 1;i <= n;i++)        scanf("%d",&x[i]);    for(int i = 1;i <= n;i++)        scanf("%d",&v[i]);    int cnt = 0;    double left = 0,right = 1e9,mid;    while(cnt++ < 123)    {        mid = (left + right) / 2;        double g = -INF,h = INF;        for(int i = 1;i <= n;i++)        {            double p = x[i] - v[i]*mid,q = x[i] + v[i]*mid;            g = max(p,g);            h = min(q,h);        }        if(g <= h)//时间太大了            right = mid;        else//            left = mid;    }    printf("%.10f/n",left);    return 0;}
由于颓废了一个假期，实在不敢打，先补了两个题，c、d待