Codeforces Round #644 (Div. 3)

A

分两种情况
设 a < b
要么 宽(a)两倍覆盖长(b), 边长为 a * 2
要么 直接长(b)覆盖宽 边长 b

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef double db;
 
const int N = 1e5 + 5;
 
int n, m, _, k;
 
int main()
{
    ios::sync_with_stdio(0); cin.tie(0);
    for (cin >> _; _; --_)
    {
        int a, b, c; cin >> a >> b;
        if (a > b) swap(a, b);
        if ((a << 1) >= b) c = (a << 1);
        else c = b;
        cout << c * c << '
';
    }
    return 0;
}

B

排序 取相邻差最小值

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef double db;
 
const int N = 50 + 5;
 
int n, m, _, k;
int s[N];
 
int main()
{
    ios::sync_with_stdio(0); cin.tie(0);
    for (cin >> _; _; --_)
    {
        cin >> n;
        rep (i, 1, n) cin >> s[i];
        sort(s + 1, s + 1 + n);
        m = 10000;
        rep (i, 2, n) m = min(m, s[i] - s[i - 1]);
        cout << m << '
';
    }
    return 0;
}

C

奇偶都为偶数对,yes
or 寻找一对 a, a + 1, 变成上面的偶数对的情况

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef double db;
 
const int N = 50 + 5;
 
int n, m, _, k;
int s[N];
 
int main()
{
    ios::sync_with_stdio(0); cin.tie(0);
    for (cin >> _; _; --_)
    {
        cin >> n; int a = 0, b = 0; 
        rep (i, 1, n)
        {
            cin >> s[i];
            if (s[i] & 1) ++a;
            else ++b;
        } 
        if (a % 2 == 0) { cout << "YES
"; continue; }
        sort(s + 1, s + 1 + n);
        int flag = 0;
        rep (i, 2, n) if (s[i] == s[i - 1] + 1) { flag = 1; break; }
        if (flag) cout << "YES
";
        else cout << "NO
";
    }
    return 0;
}

D

暴力

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef double db;
 
const int N = 50 + 5;
 
int n, m, _, k;
int s[N];
 
int main()
{
    ios::sync_with_stdio(0); cin.tie(0);
    for (cin >> _; _; --_)
    {
        cin >> n >> k;
        int ans = n;
        for (int i = 1; i <= k && i * i <= n; ++i)
            if (n % i == 0)
            {
                int a = n / i;
                if (a <= k) { ans = i; break; }
                ans = a;
            }
        cout << ans << '
';
    }
    return 0;
}

E

从最外层,一层一层判断

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef double db;
 
const int N = 50 + 5;
 
int n, m, _, k;
char s[N][N];
 
int main()
{
    ios::sync_with_stdio(0); cin.tie(0);
    for (cin >> _; _; --_)
    {
        cin >> n; bool flag = 1;
        rep (i, 1, n) cin >> s[i] + 1;
 
        per (k, n, 1)
        {
            int h = n - k + 1;
            per (i, h, 1)
                if (s[h][i] == '1' && !(h == n || i == n || s[h + 1][i] == '1' || s[h][i + 1] == '1'))
                {
                    //cout << h << '!' << i << '
';
                    flag = 0;
                    break;
                }
            
            per (i, h, 1)
                if (s[i][h] == '1' && !(h == n || i == n || s[i + 1][h] == '1' || s[i][h + 1] == '1'))
                {
                    //cout << s[i] << '
';
                    //cout << i << ' ' << h << '
';
                    flag = 0;
                    break;
                }
 
            if (flag == 0) break;
        }
 
        if (flag == 0) cout << "NO
";
        else cout << "YES
";
    }
    return 0;
}

F

dfs

#include <bits/stdc++.h>
#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef double db;
 
const int N = 10 + 5;
 
int n, m, _, k;
char s[N][N], ans[N];
int st[N];
 
bool dfs(int k)
{
    if (k == m + 1) return true;
 
    char c = '1'; int cur[N];
    rep(i, 1, n) cur[i] = st[i];
    rep(i, 1, n)
        if (s[i][k] != c)
        {
            c = ans[k] = s[i][k];
            bool flag = 1;
            rep(j, 1, n)
            {
                if (s[j][k] != c) ++st[j];
                if (st[j] > 1)
                {
                    rep(p, 1, j) st[p] = cur[p];
                    flag = 0; break;
                }
            }
 
            if (flag)
            {
                if (dfs(k + 1)) return true;
                rep(p, 1, n) st[p] = cur[p];
            }
        }
 
    return false;
}
 
 
int main()
{
    ios::sync_with_stdio(0); cin.tie(0);
    for (cin >> _; _; --_)
    {
        cin >> n >> m;
        rep(i, 1, n) cin >> s[i] + 1;
        memset(st, 0, sizeof st);
 
        ans[m + 1] = '';
        if (dfs(1)) cout << ans + 1 << '
';
        else cout << -1 << '
';
    }
    return 0;
}

G

dancing links 不会

H

先找到中位数, 然后左右移动中位数就行
用set存一下删掉的数

#define all(n) (n).begin(), (n).end()
#define se second
#define fi first
#define pb push_back
#define mp make_pair
#define sqr(n) (n)*(n)
#define rep(i,a,b) for(int i=a;i<=b;++i)
#define per(i,a,b) for(int i=a;i>=b;--i)
using namespace std;
typedef long long ll;
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef double db;
 
const int N = 60 + 5;
 
int n, m, _, k;
ll b[N];
char s[N];
 
int main()
{
    ios::sync_with_stdio(0); cin.tie(0);
    b[0] = 1;
    rep (i, 1, 61) b[i] = b[i - 1] << 1;
    for (cin >> _; _; --_)
    {
        cin >> n >> m;
        ll ans = 1ll << (m - 1); --ans;
        ll x = ans, y = ans + 1;
 
        unordered_set<ll> mp;
        rep (i, 1, n)
        {
            cin >> s + 1;
            ll res = 0;
            for (int i = 0, j = m; j; --j, ++i)
                if (s[j] == '1') res += b[i];
 
            mp.insert(res); 
            if (res == ans) 
            {
                if (y)
                {
                    ++ans, --y;
                    while (mp.count(ans)) ++ans;
                }
                else 
                {
                    --ans, --x;
                    while (mp.count(ans)) --ans;
                }
            }
            else if (res > ans) --y;
            else --x;
 
            while (x > y)
            {
                --x, ++y; --ans;
                while (mp.count(ans)) --ans; 
            }
            while (y > x + 1)
            {
                --y, ++x, ++ans;
                while (mp.count(ans)) ++ans;
            }
        }
        
        for (int i = m - 1, j = 1; j <= m; ++j, --i)
            if (ans >= b[i]) cout << 1, ans -= b[i];
            else cout << 0;
        cout << '
';
    }
    return 0;
}