目录

• A - Five Antennas
• B - Five Dishes
• C - Five Transportations
• D - Cake 123

A - Five Antennas

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;
typedef long long LL;
int a[5];
int main() {
    for (int i = 0; i < 5; i++) cin >> a[i];
    int k;
    cin >> k;
    for (int i = 0; i < 4; i++) {
        for (int j = i + 1; j < 5; j++) {
            if (a[j] - a[i] > k) {
                cout << ":(" << endl;
                return 0;
            }
        }
    }
    cout << "Yay!" << endl;
    return 0;
}

B - Five Dishes

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;
typedef long long LL;
int a[5];
int main() {
    int res = 0x3f3f3f3f, sum = 0;
    for (int i = 0; i < 5; i++) {
        cin >> a[i];
        sum += int(ceil(1.0 * a[i] / 10)) * 10;
    }
    for (int i = 0; i < 5; i++) {
        res = min(res, sum - (int(ceil(1.0 * a[i] / 10)) * 10 - a[i]));
    }
    cout << res << endl;
    return 0;
}

C - Five Transportations

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;
typedef long long LL;
LL a[5];
int main() {
    LL MIN = 0x3f3f3f3f3f3f3f3f;
    int pos = -1;
    LL n;
    cin >> n;
    for (int i = 0; i < 5; i++) {
        cin >> a[i];
        if (a[i] < MIN) {
            MIN = a[i];
            pos = i;
        }
    }
    cout << 4 + LL(ceil(1.0*n / MIN)) << endl;
    return 0;
}

D - Cake 123

给出三个数组，长度均为1e3

要求求出这三个数组任意组合中前k大的数（即前k大的ai+bj+ck)，k小于3000

直接暴力会超时，所以可以先n方求出a与b的和，然后取前k大的数，与c求和，再取前k大的数，这样复杂度为n方

#include <bits/stdc++.h>

using namespace std;

const int N = 1e7 + 5;
typedef long long LL;
int X, Y, Z, k;
LL x[N], y[N], z[N];
LL tmp[N], res[N];
int main() {
    cin >> X >> Y >> Z >> k;
    for (int i = 0; i < X; i++) cin >> x[i];
    for (int i = 0; i < Y; i++) cin >> y[i];
    for (int i = 0; i < Z; i++) cin >> z[i];
    int cnt = 0;
    for (int i = 0; i < X; i++) {
        for (int j = 0; j < Y; j++) {
            tmp[cnt++] = x[i] + y[j];
        }
    }
    sort(tmp, tmp + cnt,greater<LL>());
    cnt = 0;
    for (int i = 0; i < min(k, X * Y); i++) {
        for (int j = 0; j < Z; j++) {
            res[cnt++] = tmp[i] + z[j];
        }
    }
    sort(res, res + cnt,greater<LL>());
    for (int i = 0; i < k; i++) {
        cout << res[i] << endl;
    }
    return 0;
}