Codeforces Round #691 (Div. 2) A~C

A.

签到，因为是等概率，且红色数字两两不同，黑色数字两两不同，故只需要看第一张即可。然后就是枚举每张卡片比较上面的两个数字。

#include <iostream>
using namespace std;
int main()
{
	freopen("data.txt", "r", stdin);
	int t;
	cin >> t;
	while(t--)
	{
		int red = 0, blue = 0, equal = 0;
		string s1, s2;
		int n;
		cin >> n;
		cin >> s1;
		cin >> s2;
		for(int i = 0; i < n; i++)
		{
			if(s1[i] > s2[i]) red++;
			else if(s1[i] == s2[i]) equal++;
			else blue++;
		}
		if(red > blue) cout << "RED" << endl;
		else if(red == blue) cout << "EQUAL" << endl;
		else cout << "BLUE" << endl;
	}
	return 0;
}

B.

打表找规律。。。

#include <iostream>
using namespace std;
int main()
{
	freopen("data.txt", "r", stdin);
	int n;
	cin >> n;
	if(n & 1)
	{
		int nn = (n + 1) / 2;
		cout << 4 + (nn - 1) * 8 + 2 * (nn - 1) * (nn - 2);
	}
	else
	{
		int nn = n / 2 + 1;
		cout << nn * nn;
	}
}

C.

不妨令b > a，注意到，(gcd(a, b) = gcd(a, b - a))，故有(gcd(a + c, b + c) = gcd(a + c, b - a))。又因为(gcd(a, b, c) = gcd(a, gcd(b, c)))， 故有(gcd(a_0 + b, a_1+b,a_2+b) = gcd(a_0+b, a_1+b, a_2-a_1)=gcd(a_0+b,a_1-a_0,a_2-a_1))...

因此我们可以先把(GCD=gcd(a_1-a_0,a_2-a_1,...,a_n-a_{n - 1}))求出来，然后遍历b数组求出(gcd(a_0+b_i,GCD))输出即可。

#include <iostream>
using namespace std;
int n, m;
long long a[200005], b[200005];
long long gcd(long long a, long long b)
{
	return b ? gcd(b, a % b) : a;
}
int main()
{
	freopen("data.txt", "r", stdin);
	cin >> n >> m;
	for(int i = 1; i <= n; i++) cin >> a[i];
	for(int i = 1; i <= m; i++) cin >> b[i];
	sort(a + 1, a + n + 1);
	long long GCD = a[2] - a[1];
	for(int i = 3; i <= n; i++) GCD = gcd(GCD, a[i] - a[i - 1]);
	for(int i = 1; i <= m; i++)
	{
		cout << gcd(a[1] + b[i], GCD) << ' ';
	}
}