解题报告：Codeforces Round #364 (Div. 2) A~E

A. Cards

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

There are n cards (n is even)
in the deck. Each card has a positive integer written on it. n / 2 people will play new card game. At the beginning of the game each
player gets two cards, each card is given to exactly one player.

Find the way to distribute cards such that the sum of values written of the cards will be equal for each player. It is guaranteed that it is always possible.

Input

The first line of the input contains integer n (2 ≤ n ≤ 100) —
the number of cards in the deck. It is guaranteed that n is even.

The second line contains the sequence of n positive integers a1, a2, ..., an (1 ≤ ai ≤ 100),
where ai is
equal to the number written on the i-th card.

Output

Print n / 2 pairs of integers, the i-th
pair denote the cards that should be given to the i-th player. Each card should be given to exactly one player. Cards are numbered
in the order they appear in the input.

It is guaranteed that solution exists. If there are several correct answers, you are allowed to print any of them.

Examples

input

6
1 5 7 4 4 3

output

1 3
6 2
4 5

input

4
10 10 10 10

output

1 2
3 4

Note

In the first sample, cards are distributed in such a way that each player has the sum of numbers written on his cards equal to 8.

In the second sample, all values ai are
equal. Thus, any distribution is acceptable.

题意:

n个数分成n/2，要求每组两个数之和相等，输出下标，保证有解。

思路：

排序后依次输出最小和最大值下标即可

代码：

#include<bits/stdc++.h>

using namespace std;

int A[105];
int B[105];

bool cmp(int x,int y){
return A[x]<A[y];
}

int main()
{
int n;
while(scanf("%d",&n)==1){
int sum = 0;
for(int i=1;i<=n;i++){
B[i]=i;
scanf("%d",&A[i]);
}sort(B+1,B+n+1,cmp);

for(int i=1;i<=n/2;i++){
printf("%d %d\n",B[i],B[n-i+1]);
}

}
}

B. Cells Not Under Attack

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Vasya has the square chessboard of size n × n and m rooks.
Initially the chessboard is empty. Vasya will consequently put the rooks on the board one after another.

The cell of the field is under rook's attack, if there is at least one rook located in the same row or in the same column with this cell. If there is a rook located in the cell, this cell is also under attack.

You are given the positions of the board where Vasya will put rooks. For each rook you have to determine the number of cells which are not under attack after Vasya puts it on the board.

Input

The first line of the input contains two integers n and m (1 ≤ n ≤ 100 000, 1 ≤ m ≤ min(100 000, n2)) —
the size of the board and the number of rooks.

Each of the next m lines contains integers xi and yi (1 ≤ xi, yi ≤ n) —
the number of the row and the number of the column where Vasya will put the i-th rook. Vasya puts rooks on the board in the order
they appear in the input. It is guaranteed that any cell will contain no more than one rook.

Output

Print m integer, the i-th
of them should be equal to the number of cells that are not under attack after first irooks are put.

Examples

input

3 3
1 1
3 1
2 2

output

4 2 0

input

5 2
1 5
5 1

output

16 9

input

100000 1
300 400

output

9999800001

Note

On the picture below show the state of the board after put each of the three rooks. The cells which painted with grey color is not under the attack.



题意：

n*n棋局,依次放入M个棋子，每个棋子的攻击范围为同一直线（同行或者同列），每次放入一个棋子询问棋盘上还有多少空间不在所有棋子的攻击范围内。

分析：

初始未被覆盖的空间ans为n*n，每次放入一个棋子，如同行未放有其他棋子，则ans-=(n-x)，x为已放有棋子的列数，然后y++;列上也进行同样的操作。

代码:

#include<bits/stdc++.h>

using namespace std;

bool h[100005];
bool l[100005];
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)==2){
memset(h,0,sizeof(h));
memset(l,0,sizeof(l));
int x=0,y=0;
long long sum = 1LL*n*n;
for(int i=1,a,b;i<=m;i++){
scanf("%d%d",&a,&b);
if(!h[a]){
sum -= n-x;
h[a]=true;
y++;
}if(!l[b]){
sum -= n-y;
l[b] = true;
x++;
}

printf("%I64d ",sum);

}printf("\n");
}return 0;
}

C. They Are Everywhere

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Sergei B., the young coach of Pokemons, has found the big house which consists of n flats ordered in a row from left to right. It is
possible to enter each flat from the street. It is possible to go out from each flat. Also, each flat is connected with the flat to the left and the flat to the right. Flat number 1 is
only connected with the flat number 2 and the flat number n is
only connected with the flat number n - 1.

There is exactly one Pokemon of some type in each of these flats. Sergei B. asked residents of the house to let him enter their flats in order to catch Pokemons. After consulting the residents of the house decided to let Sergei B. enter one flat from the street,
visit several flats and then go out from some flat. But they won't let him visit the same flat more than once.

Sergei B. was very pleased, and now he wants to visit as few flats as possible in order to collect Pokemons of all types that appear in this house. Your task is to help him and determine this minimum number of flats he has to visit.

Input

The first line contains the integer n (1 ≤ n ≤ 100 000)
— the number of flats in the house.

The second line contains the row s with the length n,
it consists of uppercase and lowercase letters of English alphabet, thei-th letter equals the type of Pokemon, which is in the flat
number i.

Output

Print the minimum number of flats which Sergei B. should visit in order to catch Pokemons of all types which there are in the house.

Examples

input

3
AaA

output

2

input

7
bcAAcbc

output

3

input

6
aaBCCe

output

5

Note

In the first test Sergei B. can begin, for example, from the flat number 1 and end in the flat number 2.

In the second test Sergei B. can begin, for example, from the flat number 4 and end in the flat number 6.

In the third test Sergei B. must begin from the flat number 2 and end in the flat number 6.

题意:

一条街上有n栋房子，每个房子上有一个字母标号，同时里面有一只相同字母标号的神奇宝贝，你可以选一个房子作为入口进入，只能一直向右或向左走，询问抓到所有类型的神奇宝贝所需的走的房子的最小数量。

思路：

基础的尺取，每当选定的区间包含所有类型，记录最小长度即可

#include<bits/stdc++.h>

using namespace std;

map<char,int>M;
set<char>S;
char str[100005];
int main()
{
int n;
while(scanf("%d",&n)==1){
M.clear();S.clear();
scanf("%s",str+1);
for(int i=1;i<=n;i++){
S.insert(str[i]);
}int sum = S.size();
int s = 1 , e = 0;
int all = 0;
int ans = n;
while(++e<=n){
M[str[e]]++;
if(M[str[e]]==1){
all++;
}

if(all==sum){
while(M[str[s]]!=1){
M[str[s++]]--;
}ans = min(ans,e-s+1);
}

}printf("%d\n",ans);

}
}

D. As Fast As Possible

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

On vacations n pupils decided to go on excursion and gather all together. They need to overcome the path with the length l meters.
Each of the pupils will go with the speed equal to v1.
To get to the excursion quickly, it was decided to rent a bus, which has seats for k people (it means that it can't fit more than kpeople
at the same time) and the speed equal to v2.
In order to avoid seasick, each of the pupils want to get into the bus no more than once.

Determine the minimum time required for all n pupils to reach the place of excursion. Consider that the embarkation and disembarkation
of passengers, as well as the reversal of the bus, take place immediately and this time can be neglected.

Input

The first line of the input contains five positive integers n, l, v1, v2 and k (1 ≤ n ≤ 10 000, 1 ≤ l ≤ 109,1 ≤ v1 < v2 ≤ 109, 1 ≤ k ≤ n) —
the number of pupils, the distance from meeting to the place of excursion, the speed of each pupil, the speed of bus and the number of seats in the bus.

Output

Print the real number — the minimum time in which all pupils can reach the place of excursion. Your answer will be considered correct if its absolute or relative error won't exceed 10 - 6.

Examples

input

5 10 1 2 5

output

5.0000000000

input

3 6 1 2 1

output

4.7142857143

Note

In the first sample we should immediately put all five pupils to the bus. The speed of the bus equals 2 and the distance is equal to 10,
so the pupils will reach the place of excursion in time 10 / 2 = 5.

题意：
n个同学出去玩，总路程为l，有一辆车可有装k个人，人行走的速度为v1，车的速度为v2。询问所有人到达终点的最少时间。

思路：

这是一道初中还是小学的数学题，可以直接手推公式或者二分到达的时间，公式好像不难推但有点复杂=-=所以我直接二分了

代码：

#include<bits/stdc++.h>

const double EPS  = 1e-9;
using namespace std;
int n,l,v1,v2,k;

inline bool check(double time){
double s = time*v1;

int cnt = (n+k-1)/k;

if(s>=l){
return true;
}

double t = 1.0*(l-s)/(v2-v1);

double tt = 1.0*(v2-v1)*t/(v2+v1);

return cnt*t+(cnt-1)*tt<=time;

}

int main()
{

while(scanf("%d%d%d%d%d",&n,&l,&v1,&v2,&k)==5){
double s = 0 ,e = 1.0*l/v1;

while(s<e-EPS){
double mid = (s+e)/2.0;
//printf("mid=%.6f\n",mid);
if(check(mid)){
e = mid ;
}else {
s = mid ;
}
}

printf("%.7f\n",e);

}return 0;
}

E. Connecting Universities

time limit per test
3 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Treeland is a country in which there are n towns connected by n - 1 two-way
road such that it's possible to get from any town to any other town.

In Treeland there are 2k universities which are located in different towns.

Recently, the president signed the decree to connect universities by high-speed network.The Ministry of Education understood the decree in its own way and decided that it was enough to connect each university with another one by using a cable. Formally, the
decree will be done!

To have the maximum sum in the budget, the Ministry decided to divide universities into pairs so that the total length of the required cable will be maximum. In other words, the total distance between universities in k pairs
should be as large as possible.

Help the Ministry to find the maximum total distance. Of course, each university should be present in only one pair. Consider that all roads have the same length which is equal to 1.

Input

The first line of the input contains two integers n and k (2 ≤ n ≤ 200 000, 1 ≤ k ≤ n / 2) —
the number of towns in Treeland and the number of university pairs. Consider that towns are numbered from 1 to n.

The second line contains 2k distinct integers u1, u2, ..., u2k (1 ≤ ui ≤ n) —
indices of towns in which universities are located.

The next n - 1 line contains the description of roads. Each line contains the pair of integers xj and yj (1 ≤ xj, yj ≤ n),
which means that the j-th road connects towns xj and yj.
All of them are two-way roads. You can move from any town to any other using only these roads.

Output

Print the maximum possible sum of distances in the division of universities into k pairs.

Examples

input

7 21 5 6 21 33 24 53 7
4 34 6

output

6

input

9 33 2 1 6 5 98 93 22 7
3 4
7 64 52 1
2 8

output

9

Note

The figure below shows one of possible division into pairs in the first test. If you connect universities number 1 and 6(marked
in red) and universities number 2 and 5 (marked
in blue) by using the cable, the total distance will equal 6 which will be the maximum sum in this example.

题意：

给一棵树，树上每条边的为单位长，其中树上2*k个点带标记，将带标记的点分成k组，每组2个，询问可以得到的最大组距和。

思路：

类似于求树上两两点距和，考虑每条可以被选择的最大次数应该为min(边的两边的带标记的点数目)，最后加起来就是答案。

代码：

#include<bits/stdc++.h>

using namespace std;

vector<int>V[200005];

bool vis[200005];
bool used[200005];
int A[200005];
int n,k;
long long ans ;

void dfs(int x){
A[x] = used[x];
vis[x] = true;
for(int i=0;i<V[x].size();i++){
int& t = V[x][i];
if(!vis[t]){
dfs(t);
A[x] += A[t];
ans += min(2*k-A[t],A[t]);
}
}return ;
}

int main()
{

while(scanf("%d%d",&n,&k)==2){
for(int i=1;i<=n;i++){
V[i].clear();
}memset(used,0,sizeof(used));
memset(vis,0,sizeof(vis));
memset(A,0,sizeof(A));
for(int i=1,a;i<=2*k;i++){
scanf("%d",&a);
used[a]=true;
}ans = 0;
for(int i=1,a,b;i<n;i++){
scanf("%d%d",&a,&b);
V[a].push_back(b);
V[b].push_back(a);
}

dfs(1);
/*
for(int i=1;i<=n;i++){
printf("%d----->%d\n",i,A[i]);
}
*/
printf("%I64d\n",ans);

}return 0 ;
}