Codeforces Round #431 (Div. 2) A B C

A. Odds and Ends

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Where do odds begin, and where do they end? Where does hope emerge, and will they ever break?

Given an integer sequence a1, a2, ..., an of length
n. Decide whether it is possible to divide it into an odd number of non-empty subsegments, the each of which has an odd length and begins and ends with odd numbers.

A subsegment is a contiguous slice of the whole sequence. For example,
{3, 4, 5} and {1} are subsegments of sequence
{1, 2, 3, 4, 5, 6}, while {1, 2, 4} and
{7} are not.

Input
The first line of input contains a non-negative integer
n (1 ≤ n ≤ 100) — the length of the sequence.

The second line contains n space-separated non-negative integers
a1, a2, ..., an (0 ≤ ai ≤ 100)
— the elements of the sequence.

Output
Output "Yes" if it's possible to fulfill the requirements, and "No" otherwise.

You can output each letter in any case (upper or lower).

Examples

Input

3
1 3 5

Output

Yes

Input

5
1 0 1 5 1

Output

Yes

Input

3
4 3 1

Output

No

Input

4
3 9 9 3

Output

No

Note
In the first example, divide the sequence into 1 subsegment:
{1, 3, 5} and the requirements will be met.

In the second example, divide the sequence into 3 subsegments:
{1, 0, 1}, {5},
{1}.

In the third example, one of the subsegments must start with
4 which is an even number, thus the requirements cannot be met.

In the fourth example, the sequence can be divided into
2 subsegments: {3, 9, 9},
{3}, but this is not a valid solution because 2 is an even number.

Code:

#include <bits/stdc++.h>
using namespace std;
const int AX = 1e2+6;
int a[AX];
int main(){
int n;
cin >> n;
int even,odd;
for( int i = 0 ; i < n ; i++ ){
cin >> a[i];
(a[i] % 2) ? odd++ : even++;
}
if( a[0] % 2 == 0 || a[n-1] % 2 == 0 ) cout << "NO" << endl;
else if( !even && odd % 2 == 1 ) cout << "YES" << endl;
else if( !even && odd % 2 == 0 ) cout << "NO" <<endl;
else{
if( n % 2 == 1 ) cout << "YES" << endl;
else cout << "NO" << endl;
}
return 0;
}

B. Tell Your World

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Connect the countless points with lines, till we reach the faraway yonder.

There are n points on a coordinate plane, the
i-th of which being
(i, yi).

Determine whether it's possible to draw two parallel and non-overlapping lines, such that every point in the set lies on
exactly one of them, and each of them passes through
at least one point in the set.

Input
The first line of input contains a positive integer n (3 ≤ n ≤ 1 000) — the number of points.

The second line contains n space-separated integers
y1, y2, ..., yn ( - 109 ≤ yi ≤ 109)
— the vertical coordinates of each point.

Output
Output "Yes" (without quotes) if it's possible to fulfill the requirements, and "No" otherwise.

You can print each letter in any case (upper or lower).

Examples

Input

5
7 5 8 6 9

Output

Yes

Input

5
-1 -2 0 0 -5

Output

No

Input

5
5 4 3 2 1

Output

No

Input

5
1000000000 0 0 0 0

Output

Yes

Note
In the first example, there are five points: (1, 7),
(2, 5), (3, 8),
(4, 6) and (5, 9). It's possible to draw a line that passes through points
1, 3, 5, and another one that passes through points
2, 4 and is parallel to the first one.

In the second example, while it's possible to draw two lines that cover all points, they cannot be made parallel.

In the third example, it's impossible to satisfy both requirements at the same time.

思路：如果两条线斜率相同，只需看 1 2 3 这三个点构成的三个斜率即可判断是否全部在两条平行的直线上。

Code:

#include <bits/stdc++.h>
#define INF 0x3f3f3f
using namespace std;
const int AX = 1e3+66;
int a[AX];
int n;
bool solve( double k ){
int temp = -1;
int falg = 0;
for( int i = 2 ; i <= n ; i++ ){
if( (a[i] - a[1]) == k * (i - 1) ) continue;
falg = 1;
if( temp < 0 ) temp = i;
else{
if( (a[i] - a[temp]) != k * ( i - temp) ) {falg = 0 ; break;}
}
}
return falg == 0 ? false : true;
}

int main(){
cin >> n;
for( int i = 1 ; i <= n ; i ++ ){
cin >> a[i];
}
double k1 = (a[2]-a[1])*1.0;
double k2 = (a[3]-a[2])*1.0;
double k3 = (a[3]-a[1])*0.5;
if( solve(k1) || solve(k2) || solve(k3) ){
cout << "Yes" << endl;
}else{
cout << "No" << endl;
}
return 0;
}

A. From Y to Y

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

From beginning till end, this message has been waiting to be conveyed.

For a given unordered multiset of n lowercase English letters ("multi" means that a letter may appear more than once), we treat all letters as strings of length
1, and repeat the following operation
n - 1 times:

Remove any two elements s and
t from the set, and add their concatenation
s + t to the set.
The cost of such operation is defined to be

, where
f(s, c) denotes the number of times character
c appears in string
s.

Given a non-negative integer k, construct any valid non-empty set of no more than
100 000 letters, such that the minimum accumulative cost of the whole process is
exactly k. It can be shown that a solution always exists.

Input
The first and only line of input contains a non-negative integer
k (0 ≤ k ≤ 100 000) — the required minimum cost.

Output
Output a non-empty string of no more than 100 000 lowercase English letters — any multiset satisfying the requirements, concatenated to be a string.

Note that the printed string doesn't need to be the final concatenated string. It only needs to represent an unordered multiset of letters.

Examples

Input

12

Output

abababab

Input

3

Output

codeforces

Note
For the multiset {'a',
'b', 'a',
'b', 'a',
'b', 'a',
'b'}, one of the ways to complete the process is as follows:

{"ab",
"a", "b",
"a", "b",
"a", "b"}, with a cost of
0;
{"aba",
"b", "a",
"b", "a",
"b"}, with a cost of 1;
{"abab",
"a", "b",
"a", "b"}, with a cost of
1;
{"abab",
"ab", "a",
"b"}, with a cost of 0;
{"abab",
"aba", "b"}, with a cost of
1;
{"abab",
"abab"}, with a cost of 1;
{"abababab"}, with a cost of
8.
The total cost is 12, and it can be proved to be the minimum cost of the process.

思路：一直用一个字母累加，直到满足等于n这个条件。

Code:

#include <bits/stdc++.h>
using namespace std;
const int AX = 1e5+666;
int main(){
int n;
ios_base::sync_with_stdio(false);
cin.tie(0);
cin >> n;
if( n == 0 ) {cout << 'a' << endl;return 0;}
int ans = 0 ;
string rs = "";
for( char i = 'a' ; i <= 'z' ; i++ ){
if( ans == n ) break;
for( int j = 0 ; j + ans <= n ; j ++ ){
ans += j ;
rs += i;
}
}
cout << rs << endl;
return 0;
}