2023牛客寒假算法基础集训营1

比赛链接https://ac.nowcoder.com/acm/contest/46800

A

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

#define int long long

const int P = 1000000007;
const int MAXN = 100005;
const int INF = 0x3f3f3f3f;

void solve() {
    string a;
    cin >> a;
    int ans = -1;
    int A = 0;
    int B = 0;
    int C = 5;
    int D = 5;
    for (int i = 0; i < 10; i++) {
        if (a[i] == '1') {
            if (i % 2) {
                B++;
            } else {
                A++;
            }
        } else {
            if (i % 2) {
                D--;
            } else {
                C--;
            }
        }
        if (A > D || B > C) {
            ans = i + 1;
            break;
        }
    }
    cout << ans;
    cout << '\n';
    return;
}

signed main() {
    int T = 1;
    cin >> T;
    while (T--)
        solve();
    return 0;
}

C

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

// #define int long long

const int P = 1000000007;
const int MAXN = 100005;
const int INF = 0x3f3f3f3f;

void solve() {
    int ans = 0;
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) {
        int a;
        cin >> a;
        if (a)
            ans++;
    }
    cout << ans << '\n';
    return;
}

signed main() {
    int T = 1;
    cin >> T;
    while (T--)
        solve();
    return 0;
}

D

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

#define int long long

const int P = 1000000007;
const int MAXN = 100005;
const int INF = 0x3f3f3f3f;

double cal(int x, int y, int X, int Y, int XX, int YY) {
    double cap = abs(min(x, X) - XX) * abs(min(y, Y) - YY);
    double yuce = abs(X - XX) * abs(Y - YY);
    double GT = x * y;
    double cup = GT + yuce - cap;
    double ans = cap / cup;
    return ans;
}

void solve() {
    int x, y, X, Y;
    cin >> x >> y >> X >> Y;
    double ans = 0;
    ans = max(ans, cal(x, y, X, Y, 0, 0));
    ans = max(ans, cal(x, y, X, Y, x, 0));
    ans = max(ans, cal(x, y, X, Y, 0, y));
    ans = max(ans, cal(x, y, X, Y, x, y));
    printf("%.9lf\n", ans);
    return;
}

signed main() {
    int T = 1;
    cin >> T;
    while (T--)
        solve();
    return 0;
}

Ｈ

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

#define int long long

const int P = 1000000007;
const int MAXN = 100005;
const int INF = 0x3f3f3f3f;

void solve() {
    int n;
    cin >> n;
    int ans = 10;
    int m = n * n - 1;
    for (int i = 0; i < m; i++) {
        string a;
        cin >> a;
        for (int j = 0; j < 4; j++) {
            if (a[j] == '1')
                ans++;
            if (a[j] == '2')
                ans--;
        }
    }
    cout << ans << '\n';
    return;
}

signed main() {
    int T = 1;
    cin >> T;
    while (T--)
        solve();
    return 0;
}

Ｋ

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

#define int long long

const int P = 1000000007;
const int MAXN = 100005;
const int INF = 0x3f3f3f3f;

void solve() {
    int n, one;
    int ans = 0;
    cin >> n >> one;
    int zero = n - one;
    if (n <= 2) {
        cout << 0 << '\n';
        return;
    }
    if (n == one) {
        cout << one - 2 << '\n';
        return;
    }
    if (zero >= 2) {
        int pair = zero / 2;
        zero -= pair * 2;
        one -= min(one, pair);
    }
    ans = max(0ll, one - 1);
    cout << ans << '\n';
    return;
}

signed main() {
    int T = 1;
    // cin >> T;
    while (T--)
        solve();
    return 0;
}

Ｌ

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

#define int long long

const int P = 1000000007;
const int MAXN = 100005;
const int INF = 0x3f3f3f3f;

void solve() {

    cout << 32;
    return;
}

signed main() {
    int T = 1;
    // cin >> T;
    while (T--)
        solve();
    return 0;
}

Ｍ

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

#define int long long

const int P = 1000000007;
const int MAXN = 505;
const int INF = 0x3f3f3f3f;

double dp[MAXN][MAXN];

void solve() {
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
        dp[i][i] = 1;
        for (int j = 1; j <= m; j++) {
            for (int k = 0; k <= j; k++) {
                dp[i][j] =
                    max(dp[i][j], 1.0 * k / j + 1.0 * dp[i - 1][(j - k)]);
            }
        }
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            cout << dp[i][j] << ' ';
        }
        cout << '\n';
    }
    printf("%.9lf\n", dp[n][m]);
    return;
}

signed main() {
    int T = 1;
    // cin >> T;
    while (T--)
        solve();
    return 0;
}