Codeforces Round #622 (Div. 2)

Codeforces Round #622 (Div. 2)

http://codeforces.com/contest/1313

A - Fast Food Restaurant

#include
using namespace std;
int t,a[3];
int main()
{
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d%d",&a[0],&a[1],&a[2]);
		int ans = 0;
		sort(a,a+3);
		for(int i = 0;i < 3;++i)
			if(a[i])
				++ans,--a[i];
		if(a[0] && a[2])
			++ans,--a[0],--a[2];
		if(a[1] && a[2])
			++ans,--a[1],--a[2];
		if(a[0] && a[1])
			++ans,--a[0],--a[1];
		if(a[0] && a[1] && a[2])
			++ans;
		printf("%d\n",ans);
	}
	return 0;
}

B - Different Rules

思路：最大尽可能凑a+b，最小尽可能凑a+b+1

#include
using namespace std;
int t,n,a,b;
int main()
{
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d%d",&n,&a,&b);
		int ans1 = 0,ans2 = 0;
		if(a > b)
			swap(a,b);
		ans2 = min(a-1,n-b) + min(b-1,n-a) + min(max(0,a-1-min(a-1,n-b)),max(0,b-1-min(b-1,n-a)));
		if(a == n && b != n)
		 	ans1 = n - 1 - min(n-b-1,a-1);
		else if(a != n && b == n)
			ans1 = n - 1 - min(n-a-1,b-1);
		else if(a == n && b == n)
			ans1 = n-1;
		else
			ans1 = n - 1 - min(n-a-1,b-1) - min(n-b-1,a-1) - min(max(0,n-a-min(n-a-1,b-1)),max(0,n-b-min(n-b-1,a-1)));
		printf("%d %d\n",ans1+1,ans2+1);
	}
	return 0;
}

C1 - Skyscrapers (easy version)

C2 - Skyscrapers (hard version)

见：https://blog.csdn.net/xing_mo/article/details/104488983