Educational Codeforces Round 32 A B C

A. Local Extrema

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

You are given an array a. Some element of this array
ai is a
local minimum iff it is strictly less than both of its neighbours (that is,
ai < ai - 1 and
ai < ai + 1). Also the element can be called
local maximum iff it is strictly greater than its neighbours (that is,
ai > ai - 1 and
ai > ai + 1). Since
a1 and
an have only one neighbour each, they are neither local minima nor local maxima.

An element is called a local extremum iff it is either local maximum or local minimum. Your task is to calculate the number of local extrema in the given array.

Input
The first line contains one integer n (1 ≤ n ≤ 1000) — the number of elements in array
a.

The second line contains n integers
a1,
a2, ...,
an (1 ≤ ai ≤ 1000) — the elements
of array a.

Output
Print the number of local extrema in the given array.

Examples

Input
31 2 3
Output
0Input
41 5 2 5
Output
2

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e3+66;
int a[maxn];
int mark[maxn];
int main(){
ios_base::sync_with_stdio(false); cin.tie(0);
int n;
memset(mark,0,sizeof(mark));
cin >> n;
for( int i = 0 ; i < n ; i++ ){
cin >> a[i] ;
}
int res = 0;
for( int i = 1 ; i < n - 1 ; i++ ){
if( !mark[i] && a[i] < a[i-1] && a[i] < a[i+1] ){
res ++ ;
mark[i] = 1;
}
if( !mark[i] && a[i] > a[i-1] && a[i] > a[i+1] ){
res ++ ;
mark[i] = 1;
}
}
cout << res << endl;
return 0;
}

B. Buggy Robot

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

Ivan has a robot which is situated on an infinite grid. Initially the robot is standing in the starting cell
(0, 0). The robot can process commands. There are four types of commands it can perform:

U — move from the cell
(x, y) to (x, y + 1);

D — move from (x, y) to
(x, y - 1);
L — move from (x, y) to
(x - 1, y);
R — move from (x, y) to
(x + 1, y).
Ivan entered a sequence of n commands, and the robot processed it. After this sequence the robot ended up in the starting cell
(0, 0), but Ivan doubts that the sequence is such that after performing it correctly the robot ends up in the same cell. He thinks that some commands were ignored by robot. To acknowledge whether the robot is severely bugged, he
needs to calculate the maximum possible number of commands that were performed correctly. Help Ivan to do the calculations!

Input
The first line contains one number n — the length of sequence of commands entered by Ivan (1 ≤ n ≤ 100).

The second line contains the sequence itself — a string consisting of
n characters. Each character can be
U, D,
L or R.

Output
Print the maximum possible number of commands from the sequence the robot could perform to end up in the starting cell.

Examples

Input

4
LDUR

Output

4

Input

5
RRRUU

Output

0

Input

6
LLRRRR

Output

4

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e2+66;
char a[maxn];
map<char,int>mp;
int main(){
int n;
cin >> n;
int res = 0;
for( int i = 0 ; i < n ; i++ ){
cin >> a[i] ;
mp[a[i]] ++;
if( a[i] == 'U' || a[i] == 'D' ){
if( mp['D'] && mp['U'] ){
int tmp = min(mp['U'] , mp['D']);
res += tmp * 2;
mp['D'] -= tmp;
mp['U'] -= tmp;
}
}else if( a[i] == 'R' || a[i] == 'L' ){
if( mp['R'] && mp['L'] ){
int tmp = min(mp['R'] , mp['L']);
res += tmp * 2;
mp['R'] -= tmp;
mp['L'] -= tmp;
}
}
}
cout << res << endl;
return 0;
}

C. K-Dominant Character

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
sta
4000ndard output

You are given a string s consisting of lowercase Latin letters. Character
c is called k-dominant iff each substring of
s with length at least
k contains this character c.

You have to find minimum k such that there exists at least one
k-dominant character.

Input
The first line contains string s consisting of lowercase Latin letters (1 ≤ |s| ≤ 100000).

Output
Print one number — the minimum value of k such that there exists at least one
k-dominant character.

Examples

Input

abacaba

Output

2

Input

zzzzz

Output

1

Input

abcde

Output

3

思路：取每个字母两两相邻的最大值，然后在出现的字母中取距离最小的那个。

Code：

#include <bits/stdc++.h>
using namespace std;
int a[26];
int x[26];
int main(){
string s;
cin >> s;
int len = s.size();
memset(x,-1,sizeof(x));
memset(a,-1,sizeof(a));
for( int i = 0 ; i < len ; i++ ){
int tmp = s[i] - 'a';
if( a[tmp] == -1 ){
x[tmp] = i + 1;
a[tmp] = i; continue;
}else{
x[tmp] = max( x[tmp] , i - a[tmp] );
a[tmp] = i;
}
}
for( int i = 0 ; i < 26 ; i++ ){
x[i] = max( len - a[i] , x[i] );
}
int res = 0x3f3f3f;
for( int i = 0 ; i < 26 ; i ++ ){
if( a[i] != -1 ){
if( x[i] != -1 ) res = min(res,x[i]);
}
}
cout << res << endl;

return 0;
}