1. A - AtCoder Quiz 2

2. B - Maritozzo

3. C - Neo-lexicographic Ordering

4. D - Strange Lunchbox

5. E - Moat

A - AtCoder Quiz 2

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

int main(){
   
    ios::sync_with_stdio(0);
    int n;
    cin >> n;
    if(n>=0&&n<40)
        cout << 40 - n;
    else if(n<70)
        cout << 70 - n;
    else if(n<90)
        cout << 90 - n;
    else
        cout << "expert";
    return 0;
}

B - Maritozzo

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

int main(){
   
    ios::sync_with_stdio(0);
    string s1, s2, s3, s4;
    cin >> s1;
    cin >> s2;
    cin >> s3;
    cin >> s4;
    int n = s4.length();
    for (int i = 0; i < n;i++){
   
        int m = s4[i] - '0';
        if(m==1)
            cout << s1;
        else if(m==2)
            cout << s2;
        else if(m==3)
            cout << s3;
    }
        return 0;
}

C - Neo-lexicographic Ordering

#include<bits/stdc++.h>
using namespace std;
int val[30];
string s;
string a[50005];
bool cmp(string &x,string &y){
   
    int mn = min(x.size(), y.size());
    for (int i = 0; i < mn;i++){
   
        if(x[i]==y[i])
            continue;
        return val[x[i] - 'a'] < val[y[i] - 'a'];
    }
    return x.size() < y.size();
}
int main(){
   
    ios::sync_with_stdio(0);
    cin >> s;
    for (int i = 0; i < 26;i++){
   
        val[s[i] - 'a'] = i;
    }
    int n;
    cin >> n;
    for (int i = 0; i < n;i++){
   
        cin >> a[i];
    }
    sort(a, a + n, cmp);
    for (int i = 0; i < n;i++){
   
        cout << a[i] << "\n";
    }
        return 0;
}

D - Strange Lunchbox

#include<bits/stdc++.h>
using namespace std;
void solve(){
   
    int n;
    cin>>n;
    int x,y;
    cin>>x>>y;
    vector<pair<int,int>>v(n);
    for(auto & p : v){
   
        cin>>p.first>>p.second;
    }
    int INF = 1e6;
    vector<vector<int>>dp(x+1, vector<int>(y+1, INF));
    dp[0][0] = 0;
    for(pair<int,int> p : v){
   
        for(int i = x; i >=0; i--){
   
            for(int j = y; j >=0; j--){
   
                int f = min(p.first + i, x);
                int s = min(p.second + j, y);
                dp[f][s] = min(dp[i][j]+1, dp[f][s]);
            }
        }
    }
 
    if(dp[x][y] >= INF){
   
        cout<<-1;
    }
    else{
   
        cout<<dp[x][y];
    }
}
int main() {
   
    // insert code here...
    solve();
    return 0;
}

E - Moat

#include<bits/stdc++.h>
using namespace std;
#define rep(i, x, y) for (int i = x; i <= y; i++)
#define dec(i, x, y) for (int i = x; i >= y; i--)
#define ll long long 
#define INF 0x3f3f3f3f

char *p1, *p2,buf[100000];
#define nc() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2) ? EOF : *p1++)
int read(){
   
    int x = 0, f = 1;
    char ch = nc();
    while (ch<48|ch>57){
   
        if(ch=='-')
            f = -1;
        ch = nc();
    }
    while(ch>=48&&ch<=57){
   
        x = x * 10 + ch - 48;
        ch = nc();
    }
    return x * f;
}

const int maxn = 105;
int n = 4, p;
int a[maxn][maxn], b[maxn][maxn], vis[maxn][maxn];
int dx[] = {
   0, 1, -1, 0};
int dy[] = {
   1, 0, 0, -1};
void dfs(int x,int y){
   
    if(x<1||y<1||x>4||y>4)
        return;
    if(b[x][y]==0||vis[x][y])
        return;
    vis[x][y] = 1;
    for (int i = 0; i < 4;i++){
   
        dfs(x + dx[i], y + dy[i]);
    }
}
bool dfs2(int x,int y){
   
    if(x<1||y<1||x>4||y>4)
        return 1;
    if(b[x][y]==1||vis[x][y])
        return 0;
    vis[x][y] = 1;
    bool ret = 0;
    for (int i = 0; i < 4;i++){
   
        ret |= dfs2(x + dx[i], y + dy[i]);
    }
    return ret;
}
int main(){
   
    ios::sync_with_stdio(0);
    n = 4;
    rep(i, 1, n) 
        rep(j, 1, n)
            a[i][j] = read();
    int cnt = 0;
    rep(val, 1, (1 << 16) - 1){
   
        int ok = 1, color = 0;
        rep(j, 0, 15) b[j / 4 + 1][j % 4 + 1] = min(1, ((val) & (1 << j)));
        rep(i, 1, 4) rep(j, 1, 4) vis[i][j] = 0;
        rep(i, 1, 4) rep(j, 1, 4) if (ok){
   
            if(b[i][j]==0&&!vis[i][j])
                ok = dfs2(i, j);
            if(a[i][j]&&(b[i][j]==0))
                ok = 0;
            if(b[i][j]==1){
   
                if(color&&!vis[i][j])
                    ok = 0;
                else if(!color&&!vis[i][j]){
   
                    dfs(i, j);
                    color = 1;
                }
            }
        }
        cnt += ok;
    }
    cout << cnt << endl;
    return 0;
}