Codeforces Round #768 (Div. 2)

A

#include <bits/stdc++.h>
using namespace std;
int a[10010], b[10010];
int main() {
    int T;
    cin >> T;
    while (T -- ) {
        int n;
        cin >> n;
        for (int i = 1; i <= n; i ++ )  cin >> a[i];
        for (int i = 1; i <= n; i ++ )  cin >> b[i];
        for (int i = 1; i <= n; i ++ ) 
            if (a[i] < b[i])    
                swap(a[i], b[i]);
        sort(a + 1, a + 1 + n);
        sort(b + 1, b + 1 + n);
        cout << a[n] * b[n] << endl;
    }
    return 0;
}

B
贪心，由于最后一个数字无法被改变，所以可以将数组反转后再从前往后贪心

#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int T, n;
int a[N];
int main() {
    cin >> T;
    while (T -- ) {
        cin >> n;
        for (int i = 1; i <= n; i ++ )  cin >> a[i];
        for (int i = 1; i <= n; i ++ )
            a[i] == a[n] ? a[i] = 1 : a[i] = 0;
        int res = 0, pos = 1;
        reverse(a + 1, a + 1 + n);
        while (pos < n) {
            if (a[pos + 1] == a[1]) {
                pos ++;
                continue;
            }
            pos *= 2;
            res ++;
        }
        cout << res << endl;
    }
    return 0;
}

C
特殊处理\(m=0和m=n-1\)的情况
首先考虑\(\sum\limits_1^{n/2}{a_i\&{b_i}}=0\)的方案
之后在考虑\(a_i \& b_i == m\)

#include <bits/stdc++.h>
#define PII pair<int, int>
using namespace std;
int T, n, m;
int main() {
    scanf("%d", &T);
    while (T -- ) {
        scanf("%d%d", &n, &m);
        int s = n;
        vector<PII> v, ans;
        n --;
        while (n >= 0) {
            int l, r;
            int x = 1;
            while (true) {
                if (x * 2 >= n) {
                    int len = n - x + 1;
                    l = x - len;
                    r = n;
                    n = x - len - 1;
                    v.push_back({l, r});
                    break;
                }
                else x *= 2;
            }
        }
        reverse(v.begin(), v.end());
        for (auto &[x, y] : v) {
            int l = x, r = y;
            while (l < r) {
                ans.push_back({l, r});
                l ++, r --;
            }
        }
        n = s;
        if (m == 0) {
            for (auto &[x, y] : ans)
                printf("%d %d\n", x, y);
            continue;
        }
        if (m == n - 1 && n > 4) {
            ans[0] = {n - 2, n - 1};
            ans[1] = {1, n - 3};
            ans[2] = {0, 2};
            for (auto &[x, y] : ans)
                printf("%d %d\n", x, y);
            continue;
        }
        int pos = -1;
        for (int i = 0; i < ans.size(); i ++ ) 
            if (ans[i].first == m || ans[i].second == m) {
                if (ans[i].first != m)  swap(ans[i].first, ans[i].second);
                pos = i;
            }
        bool flag = 0;
        for (int i = 0; i < ans.size(); i ++ ) {
            if (i == pos)   continue;
            if (((ans[i].first & ans[pos].first) == m) && ((ans[i].second & ans[pos].second) == 0)) {
                flag = true;
                swap(ans[i].second, ans[pos].first);
                break;
            }
            if (((ans[i].second & ans[pos].first) == m) && ((ans[i].first & ans[pos].second) == 0)) {
                flag = true;
                swap(ans[i].first, ans[pos].first);
                break;
            }
        }
        if (!flag)  puts("-1");
        else {
            for (auto &[x, y] : ans)
                printf("%d %d\n", x, y);
        }
    }
    return 0;
}