Codeforces Round #768 (Div. 2) A ~ C

Codeforces Round #768 (Div. 2) A ~ C https://codeforces.com/contest/1631/problem/A

题意：给定两个数组，可以选择任意下标i，交换ai和bi，使得a数组最大值 * b数组最大值最小。
题解：将ai > bi 的移动到a数组，小于的放在b数组

const int N = 110;
int a[N], b[N];

int main()
{
	int t;
	cin >> t;
	while (t--)
	{
		int n;
		cin >> n;
		int m1 = 0, m2 = 0;
		for (int i = 0; i < n; i++)
		{
			cin >> a[i];
		}
		for (int i = 0; i < n; i++)
		{
			cin >> b[i];
		}
		for (int i = 0; i < n; i++)
		{
			if (a[i] < b[i]) swap(a[i], b[i]);
		}
		sort(a, a + n);
		sort(b, b + n);
		m1 = a[n - 1], m2 = b[n - 1];
		cout << m1 * m2 << endl;
	}
	return 0;
}

https://codeforces.com/contest/1631/problem/B

题意：每次可以选定2k长度的数组，并将后k段复制到前k段去，问数组所有元素相同时的最小 *** 作次数。
题解：贪心，每次从后找到最长的相同段，然后倍增，直到大于数组长度。

const int N = 2e5 + 10;
int a[N], b[N];

int main()
{
	int t;
	cin >> t;
	while (t--)
	{
        int n;
        cin >> n;
        for (int i = 0; i < n; i++) cin >> a[i];
        int l = 0;
        for (int i = n - 1; i >= 0; i--)
        {
            if (a[i] == a[n - 1]) l++;
            else break;
        }
        int ans = 0;
        while (l < n)
        {
            l *= 2;
            for (int i = n - l - 1; i >= 0; i--)
            {
                if (a[i] == a[n - 1])
                {
                    l++;
                }
                else break;
            }
            ans++;
        }
        cout << ans << endl;
	}
	return 0;
}

https://codeforces.com/contest/1631/problem/C

题意：给定n和k，从0到n-1中构造n/2对，使得

其中，n为2的指数倍。k为0到n-1
题解：每次0和n-1配对，1和n-2，2和n-3…可以构造出0。对于0到n-1，由于n-1和k可以构造出k，原本和k配对的n-1-k和0配对任然为0。对于n-1，在配一个1和3即可。但需要特判4。

int main()
{
	int t;
	cin >> t;
	while (t--)
	{
		int n, k;
		cin >> n >> k;
		if (n == 4)
		{
			if (k == 3) cout << -1 << endl;
			else if (k == 0)
			{
				cout << 0 << ' ' << n - 1 << endl;
				cout << n - 2 << ' ' << n - 3 << endl;
			}
			else
			{
				cout << n - 1 << ' ' << k << endl;
				cout << 0 << ' ' << n - 1 - k << endl;
			}
		}
		else if (k > 0 && k < n - 1)
		{
			cout << n - 1 << ' ' << k << endl;
			cout << 0 << ' ' << n - 1 - k << endl;
			for (int i = 1, j = n - 2; i < j; i++, j--)
			{
				if(i != k && j != k) cout << i << ' ' << j << endl;
			}
 		}
		else if (k == 0)
		{
			for (int i = 0, j = n - 1; i < j; i++, j--)
			{
				cout << i << ' ' << j << endl;
			}
		}
		else
		{
			cout << n - 1 << ' ' << n - 2 << endl;
			cout << 0 << ' ' << n - 3 << endl;
			cout << 1 << ' ' << 3 << endl;
			cout << 2 << ' ' << n - 4 << endl;
			for (int i = 4, j = n - 5; i < j; i++, j--)
			{
				cout << i << ' ' << j << endl;
			}
		}
	}
	return 0;
}