n=int(input())
arr=[]
for i in range(n):arr.append(input().strip().replace('x',''))
for i in range(n):
    cnt=arr[i].count('y')
    arr[i]=arr[i].replace('y','')
    arr[i]+=cnt*'y'

arr=sorted(arr,key=lambda x:(-len(x),x))
print('\n'.join(arr))

Michael's solution:

#include "bits/stdc++.h"
using namespace std;

string pi = "31415926535897932384626433832... (truncated copy-paste from internet)";

void solve() {
    string s;
    cin >> s;
    for (int i = 0; i < s.length(); i++)
        if (s[i] != pi[i]) {
            cout << i << '\n';
            return;
        }
    cout << s.length() << '\n';
}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    int t;
    cin >> t;
    while (t--) solve();
}

n=int(input())
b=int(input())
x=int(input())
y=int(input())

s=4*(n-1)*y+(n-2)*(n-2)*x

print(["Walter will like this lab","Sorry Mr. Fring, no can do"][s>b])

Michael did this one, but it looks like pain...

#include "bits/stdc++.h"
using namespace std;

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    string s;
    cin >> s;
    int a = 0, b = 0;
    char cur = s[0];
    int len = 1;
    for (int i = 1; i < s.length(); i++) {
        if (cur != s[i]) {
            if (len >= 3) {
                if (cur == '<')
                    a -= 20 + (len - 3) * 5;
                else if (cur == '>')
                    a += 20 + (len - 3) * 5;
                else if (cur == 'v')
                    b -= 20 + (len - 3) * 5;
                else if (cur == '^')
                    b += 20 + (len - 3) * 5;
            } else {
                if (cur == '<')
                    a -= 2 * len;
                else if (cur == '>')
                    a += 2 * len;
                else if (cur == 'v')
                    b -= 2 * len;
                else if (cur == '^')
                    b += 2 * len;
            }
            len = 1;
            cur = s[i];
        } else {
            len++;
        }
    }
    if (len >= 3) {
        if (cur == '<')
            a -= 20 + (len - 3) * 5;
        else if (cur == '>')
            a += 20 + (len - 3) * 5;
        else if (cur == 'v')
            b -= 20 + (len - 3) * 5;
        else if (cur == '^')
            b += 20 + (len - 3) * 5;
    } else {
        if (cur == '<')
            a -= 2 * len;
        else if (cur == '>')
            a += 2 * len;
        else if (cur == 'v')
            b -= 2 * len;
        else if (cur == '^')
            b += 2 * len;
    }
    cout << "(" << a << "," << b << ")" << '\n';
}

Explanation coming soon.........

n=int(input())
a=[0]+[int(x)for x in input().split()]
b=0
for i in range(1,n+1):
    b+=max(a[i]-a[i-1],0)
print(b)

DFS through the tree, keeping track of time. Keep track of the tree with max time and then size as tiebreaker.

#include <stdio.h>
#include <stdlib.h>
#include <assert.h>

#include <vector>
#include <algorithm>
using namespace std;

int lastlay=-1;
int lastnode=-1;

vector<vector<int>> adj(1000);
void dfs(int v,int p,int lay){
        if(lay>lastlay)lastlay=lay,lastnode=v;
        else if(lay==lastlay&&v>lastnode)lastnode=v;
        for (auto&x:adj[v])if(x!=p)dfs(x,v,lay+1);
}

int main()
{
        int r, n; scanf("%d %d", &r, &n);
        r--;
        
        for (int i=0;i<n;i++){int u,v;scanf("%d %d",&u,&v);
                u--;v--;
                adj[u].push_back(v);
                adj[v].push_back(u);
        }
        for (auto&v:adj)sort(v.begin(),v.end());
        dfs(r,r,0);
        printf("%d\n",lastnode+1);
}

Michael did this one, so my explanation might not be great. Just brute force while keeping track of visited letters in O(3^26) time. Since the testcases are random, the average time is a lot less, so brute force works.

#include "bits/stdc++.h"
using namespace std;

int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};

int n, m;
vector<string> v;

int ans = 0;

void dfs(int x, int y, vector<int> vis, int done) {
    ans = max(ans, done);
    for (int i = 0; i < 4; i++) {
        int nx = x + dx[i], ny = y + dy[i];
        if (nx < 0 || ny < 0 || nx >= n || ny >= m) continue;
        if (vis[v[nx][ny] - 'A']) continue;
        vector<int> vis2 = vis;
        vis2[v[nx][ny] - 'A'] = true;
        dfs(nx, ny, vis2, done + 1);
    }
}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    cin >> n >> m;
    for (int i = 0; i < n; i++) {
        string s;
        cin >> s;
        v.push_back(s);
    }

    vector<int> vis(27, false);
    vis[v[0][0] - 'A'] = true;
    dfs(0, 0, vis, 1);
    cout << ans << '\n';
}

Use dp: dp[i][j] represents # of ways to create a parenthesis sequence up to index i with j extra left parenthesis (

If you are on a (: Add dp[i][j] to dp[i+1][j+1]

If you are on a ): Add dp[i][j] to dp[i+1][j-1] if j>0

If you are on a B's turn or a ?: Do both

#include "bits/stdc++.h"
using namespace std;

const int MOD = 1e9 + 7;
int add(int a, int b) { return (a + b) % MOD; }

int dp[4001][4001];

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    int n;
    string s;
    cin >> n >> s;

    if (s[0] == ')') {
        cout << 0 << '\n';
        return 0;
    }

    memset(dp, 0, sizeof dp);

    dp[0][0] = 1;

    for (int i = 0; i < n * 2; i++) {
        for (int j = 0; j < n * 2; j++) {
            if (i & 1) {
                if (j > 0)
                    dp[i + 1][j - 1] = add(dp[i + 1][j - 1], dp[i][j]);
                dp[i + 1][j + 1] = add(dp[i + 1][j + 1], dp[i][j]);
            } else {
                if (s[i / 2] == ')') {
                    if (j > 0)
                        dp[i + 1][j - 1] = add(dp[i + 1][j - 1], dp[i][j]);
                } else if (s[i / 2] == '(')
                    dp[i + 1][j + 1] = add(dp[i + 1][j + 1], dp[i][j]);
                else if (s[i / 2] == '?') {
                    if (j > 0)
                        dp[i + 1][j - 1] = add(dp[i + 1][j - 1], dp[i][j]);
                    dp[i + 1][j + 1] = add(dp[i + 1][j + 1], dp[i][j]);
                } else
                    assert(false);
            }
        }
    }

    cout << dp[2 * n][0] << '\n';
}