AtCoder Beginner Contest 330

1|0A - Counting Passes

#include <bits/stdc++.h>

using namespace std;

#define int long long
using vi = vector<int>;
using i32 = int32_t;
using pii = pair<int, int>;
#define mp make_pair
const int inf = 1e9;


i32 main() {
    int n , l;
    cin >> n >> l;
    int res = 0;
    for( int i = 1 , x ; i <= n ; i ++ )
        cin >> x , res += x >= l;
    cout << res << "\n";
    return 0;
}

2|0B - Minimize Abs 1

#include <bits/stdc++.h>

using namespace std;

#define int long long
using vi = vector<int>;
using i32 = int32_t;
using pii = pair<int, int>;
#define mp make_pair
const int inf = 1e9;


i32 main() {
    int n, l, r;
    cin >> n >> l >> r;
    for (int a, i = 1; i <= n; i++) {
        cin >> a;
        if (l <= a and a <= r) cout << a << " ";
        else if( a > r ) cout << r << " ";
        else cout << l << " ";
    }
    cout << "\n";
    return 0;
}

3|0C - Minimize Abs 2

预处理所有的平方数，然后枚举加二分

#include <bits/stdc++.h>

using namespace std;

#define int long long
using vi = vector<int>;
using i32 = int32_t;
using pii = pair<int, int>;
#define mp make_pair
const int inf = 1e9, INF = 1e18;


i32 main() {
    int n;
    cin >> n;
    vi a;
    for (int i = 0, x;; i++) {
        x = i * i, a.push_back(x);
        if (x > n) break;
    }
    int res = INF;
    for (int y; const auto &x: a) {
        if (n - x <= 0) res = min(res, x - n);
        else {
            y = lower_bound(a.begin(), a.end(), n - x) - a.begin();
            res = min(res, abs(x + a[y] - n));
            if (y > 1 ) res = min(res, abs(x + a[y - 1] - n));
        }
    }
    cout << res << "\n";
    return 0;
}

4|0D - Counting Ls

直接统计每行每列o的数量，然后枚举一下中心点，计算出两边两个的点的选择方式

#include <bits/stdc++.h>

using namespace std;

#define int long long
using vi = vector<int>;
using i32 = int32_t;
using pii = pair<int, int>;
#define mp make_pair
const int inf = 1e9, INF = 1e18;


i32 main() {
    int n;
    cin >> n;
    vector<string> g(n);
    for (auto &i: g) cin >> i;
    vi cntX(n), cntY(n);
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++) {
            if (g[i][j] == 'x') continue;
            cntX[i]++, cntY[j]++;
        }
    int res = 0;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++) {
            if (g[i][j] == 'o') res += (cntX[i] - 1) * (cntY[j] - 1);
        }
    cout << res << "\n";
    return 0;
}

5|0E - Mex and Update

用set维护当前序列中哪些数字没有即可，注意的是所有大于n的数字都可以等效为n

#include <bits/stdc++.h>

using namespace std;

#define int long long
using vi = vector<int>;
using i32 = int32_t;
using pii = pair<int, int>;
#define mp make_pair
const int inf = 1e9, INF = 1e18;


i32 main() {
    int n, q;
    cin >> n >> q;
    vector<int> cnt(n + 1), a(n);
    for (auto &i: a)
        cin >> i, i = min(i, n), cnt[i]++;
    set<int> mex;
    for (int i = 0; i <= n; i++)
        if (cnt[i] == 0) mex.insert(i);
    for (int i, x; q; q--) {
        cin >> i >> x, i--, x = min(x, n);
        cnt[a[i]]--;
        if (cnt[a[i]] == 0) mex.insert(a[i]);
        cnt[x]++;
        if (cnt[x] == 1) mex.erase(x);
        a[i] = x;
        cout << *mex.begin() << "\n";
    }

    return 0;
}