# Codeforces Global Round 17

## A. Anti Light's Cell Guessing

### analysis:

Just consider the case of 0.

### code:

#include <bits/stdc++.h>
using namespace std;
//#pragma GCC optimize(2)
#define close(); 	ios::sync_with_stdio(false);
#define endl '\n'
#define rep(i, l, r) for(int i = l; i <= r; i++)
#define dwn(i, r, l) for(int i = r; i >= l; i--)
typedef long long LL;
const int N = 3e5+100;

void solve()
{
int a, b; cin >> a >> b;
if(a > b) swap(a, b);
if(a == 1 && b == 1) cout << 0 << endl;
else if(a == 1) cout << 1 << endl;
else cout << 2 << endl;
}

int main()
{
int T; cin >> T;
while(T--) solve();
// system("pause");
}


## B. Kalindrome Array

### analysis:

Remember that the string is s and the length is m

The discussion is divided into the following situations:

1. It is a palindrome string and outputs yes
2. If it is not a palindrome string, there must be a pair of \ (s_i!=s_j, i = m+1-j \)
• Consider deleting all the characters of the \ (s_i \) class to see if it is a palindrome string again
• Consider deleting all characters of the \ (s_j \) class to see if it is a palindrome string again

Why delete all its class characters? It seems a little different from the title stem. In fact, even if it does not delete its class characters, it is the matching itself. It is equivalent to judging whether it is a palindrome string in the case of complete deletion

### code:

#include <bits/stdc++.h>
using namespace std;
//#pragma GCC optimize(2)
#define close(); 	ios::sync_with_stdio(false);
#define endl '\n'
#define rep(i, l, r) for(int i = l; i <= r; i++)
#define dwn(i, r, l) for(int i = r; i >= l; i--)
typedef long long LL;
const int N = 3e5+100;

int a[N];

bool check(int n, int x)
{
int l = 1, r = n;
while(l < r)
{
while(l < r && a[l] == a[x]) l++;
if(l >= r) break;
while(l < r && a[r] == a[x]) r--;
if(l >= r) break;
if(a[l] == a[r]) l++, r--;
else if(a[l] != a[r]) return 0;
}
return 1;
}

void solve()
{
int n; cin >> n;
rep(i, 1, n) cin >> a[i];
int l = 1, r = n;
int u, v; u = v = 0;
while(l < r)
{
if(a[l] == a[r]) l++, r--;
else
{
u = l;
v = r;
break;
}
}
bool f = 0;
if(u == v) f = 1;
else
{
f |= check(n, u);
f |= check(n, v);
}
if(f) cout << "YES\n";
else cout << "NO\n";
}

int main()
{
close();
int T; cin >> T;
while(T--) solve();
// system("pause");
}


## C. Keshi Is Throwing a Party

### analysis:

At first, I thought about it for a while. I regretted my greed and found that I couldn't do it

Then I found that it was a useful algorithm

When we determine the number of people we want to check, we directly look for it greedily. If we don't understand, read the code

Time complexity \ (O(n\log n) \)

### code:

#include <bits/stdc++.h>
using namespace std;
//#pragma GCC optimize(2)
#define close(); 	ios::sync_with_stdio(false);
#define endl '\n'
#define rep(i, l, r) for(int i = l; i <= r; i++)
#define dwn(i, r, l) for(int i = r; i >= l; i--)
typedef long long LL;
const int N = 3e5+100;

int a[N], b[N];
int n;
bool check(int x)
{
int cnt = 0;
rep(i, 1, n)
{
if(a[i] >= x-cnt-1 && b[i] >= cnt)
{
cnt++;
}
if(cnt == x) return 1;
}
return 0;
}

void solve()
{
cin >> n;
rep(i, 1, n) cin >> a[i] >> b[i];

int l = 1, r = n;
while(l < r)
{
if(l == r-1)
{
if(check(r)) l = r;
else r = l;
break;
}
int mid = l+r>>1;
if(check(mid)) l = mid;
else r = mid-1;
}
cout << l << endl;
}

int main()
{
close();
int T; cin >> T;
while(T--) solve();
// system("pause");
}


## D. Not Quite Lee

### analysis:

Consider an array \ (c \) with \ (k \) elements, and how to judge whether it is good.

We find that for each \ (c_i\in c \), the sum of consecutive sequences of integers is \ (\ frac{c_i(c_i-1)}{2}+x_ic_i \), where \ (x_i \) is any integer.

Therefore, we can convert whether an array c is good or not into this equation:

$0 = \sum_{i=1}^k \frac{c_i(c_i-1)}{2}+x_ic_i$

By observing this formula, we find that we are very familiar with it. We might as well define \ (t \) as the following formula:

$t = \sum_{i=1}^k x_ic_i$

According to peishu theorem, \ (GCD (c_1, c_2,..., c_k) | t \)

Record \ (s = \ sum_{I = 1} ^ k \ frac {c_i (c_i-1)} {2}, g = GCD (c_1, c_2,..., c_k) \)

So the problem of determining whether an array c is good is transformed into: whether g can divide s

Consider when there are odd numbers for array c:

1. If there are odd and even numbers in array c, then \ (g=1 \) has a solution
2. If there are only odd numbers in array c, let the midpoint of all continuous sequences be placed at zero, s = 0, there is a solution

So next, we only need to consider the case that there are only even numbers in array c

It is observed that if the odd number y is divided by the even number x, then \ (y|\frac{x}{2} \); If even y is divided by even x, then \ (\ frac{y}{2}|\frac{x}{2} \)

Considering that enumeration g is time infeasible, we try to establish a 1-1 mapping to reduce the divisor we need to enumerate

For \ (g|s \), obviously s is an even number and G is also an even number (the odd number has been calculated), then \ (\ frac {g} {2^l} \ frac {s} {2^l} \). Where $2^l | g \space \$and \ (\ space2^{l+1} \not | g \), that is to say \ (2^l \) is the largest integral power factor of 2 of G

Note: odd numbers are handled in advance, so \ (l\le 1 \)

So we might as well map g to \ (2^l \), which is obviously a 1-1 mapping. The specific operations are as follows:

For all elements in array c that can be divided by \ (2^l \) but cannot be divided by $2^{l+1}$, put them into set A

For all elements in array c that can be divided by \ (2^{l+1} \), put them into set B

Obviously, for any combination containing an even number of elements in set A, the mapping of g is \ (2^l \)

Why not include an odd number of elements in set A? If you look at A single element, you might as well write it as \ (e = k2^l \), where k is an odd number. Then \ (\ frac{e(e-1)}{2} = k2^{l-1}(k2^l-1) \), this is A number with the largest integral power factor of 2 as \ (2^{l-1} \). Obviously, the integral power factor of the largest 2 of s is also \ (2^{l-1} \)

But if it is an even number of set A elements, this problem can be solved

Let the number of elements in array c that can be divided by \ (2^l \) but cannot be divided by $2^{l+1}$be \ (a \); The number of all elements in array c that can be divided by \ (2^{l+1} \) is \ (b \)

Then the answer to \ (2^l \) is \ (2^{a-1}\cdot 2^b \)

Just enumerate \ (l \) statistical answers

The total time complexity is \ (O(n\log(1e9)) \)

### code:

#include <bits/stdc++.h>
using namespace std;
//#pragma GCC optimize(2)
#define close(); 	ios::sync_with_stdio(false);
#define endl '\n'
#define rep(i, l, r) for(int i = l; i <= r; i++)
#define dwn(i, r, l) for(int i = r; i >= l; i--)
typedef long long LL;
const int N = 3e5+100;
const LL p = 1e9+7;
LL qpow(LL a, LL b)
{
LL rev = 1;
while(b){
if(b&1) rev = rev * a % p;
a = a * a % p;
b>>=1;
}
return rev;
}

LL a[N];

int main()
{
// close();

int n; cin >> n;
rep(i, 1, n) cin >> a[i];
LL ans = 0;
LL even = 0;
rep(i, 1, n)
{
if(a[i] & 1) ; else even++;
}
ans = (qpow(2, n) - qpow(2,even)) % p;
rep(i, 1, 31)
{
LL x = 1ll<<i;
LL o, e; o = e = 0;
rep(j, 1, n)
{
if(a[j] % x == 0 && a[j] % (x<<1) != 0) o++;
else if(a[j] % (x<<1) == 0) e++;
}

if(o > 0) (ans += ((qpow(2, o-1)-1+p)%p) * ((qpow(2, e))%p) ) %= p;
}

cout << ans << endl;

// system("pause");
}


## E. AmShZ and G.O.A.T.

Catch up with your homework and make it up when you are free

Posted on Wed, 24 Nov 2021 10:47:24 -0500 by wtfsmd