Educational Codeforces Round 120 (Rated for Div. 2) A ~ C

Posted 2023-02-02 Ja_King_ZH

https://codeforces.com/contest/1622/problem/A

题意：给定三个正整数，将其中一个数分为两个数，判断能否构成长方形。
题解：枚举每种情况就行。

int f[4];
bool cmp(int a, int b)

	return a > b;

int main()

	int t;
	cin >> t;
	while (t--)
	
		cin >> f[0] >> f[1] >> f[2];
		sort(f, f + 3,cmp);
		int a = f[0], b = f[1], c = f[2];
 
		if (a == b)
		
			if (c % 2 == 0)cout << "YES" << endl;
			else cout << "NO" << endl;
		
		else
		
			if (a == (b + c))cout << "YES" << endl;
			else
			
				if (b == c)
					if (a % 2 == 0)cout << "YES" << endl;
					else cout << "NO" << endl;
				else cout << "NO" << endl;
			
		
	
	return 0;

https://codeforces.com/contest/1622/problem/B

题意：给定一个数组q，和相同长度的01串，要求1对应的数字必须大于0对应的数字，构造一个数组p，求

题解：暴力求解即可，首先看有多少个0和1，由于1对应的数都大于0对应的数，所以直接求解原数字排序后的下标，依次按下标填入数据。

const int N = 2e5 + 10;
typedef pair<int, int>PII;
int f[N],ans[N];
 
int main()

	int t;
	cin >> t;
	while (t--)
	
		int n;
		cin >> n;
		for (int i = 0; i < n; i++) cin >> f[i];
		string res;
		cin >> res;
		int cnt1 = 0, cnt2 = 0;
		vector<PII>v1, v2;
		for (int i = 0; i < res.size(); i++)
		
			if (res[i] == '0')cnt1++, v1.push_back( f[i], i );
			else cnt2++, v2.push_back( f[i], i );
		
		sort(v1.begin(), v1.end());
		sort(v2.begin(), v2.end());
		for (int i = 1; i <= cnt1; i++)
		
			ans[v1[i - 1].second] = i;
		
		for (int i = 1; i <= cnt2; i++)
		
			ans[v2[i - 1].second] = i + cnt1;
		
		for (int i = 0; i < n; i++) cout << ans[i] << ' ';
		cout << endl;
		for (int i = 0; i < n; i++) ans[i] = 0;
	
	return 0;

https://codeforces.com/contest/1622/problem/C

题意：给定一个数组，每次可以将某个数-1或者选择两个数，将一个数赋值给另一个数，求解数组的和小于k的最小操作次数。
题解：二分+贪心。首先排序，假设数组长度为n，那么将后n-1个数全部赋值为第一个数，设该操作次数为b，即n-1。剩下的就是将a[0] 减少的次数，设为a，当a<0时，需要改变a，b的值。然后比较b和f[j+1]-f[0]+a的大小（即减少一次赋值，多一次f[0]–的操作，哪个可以将数组的值减少的更多）

const int N = 2e5 + 10;
long long f[N];
long long n, k, sum;
bool check(long long x)

	if (sum - x <= k) return true;
	long long a, b, j;
	j = 0;
	b = n - 1;
	a = x - b;
	if (a < 0) b += a, a = 0, j = n - b - 1;
	while (1)
	
		if (b > 0 && b > (f[j + 1] - f[0] + a)) j++, b--, a++;
		else break;
	
	long long res = 0, cnm = sum;
	for (int i = n - 1; i > (n - 1) - b; i--)
	
		cnm = cnm - (f[i] - f[0] + a);
		if (cnm <= k) return true;//防止爆longlong.....巨坑！！！
	
	cnm -= a;
	if (cnm <= k) return true;
	return false;

signed main()

	int t;
	cin >> t;
	while (t--)
	
		cin >> n >> k;
		sum = 0;
		for (int i = 0; i < n; i++) cin >> f[i], sum += f[i];
		sort(f, f + n);
		long long l = 0, r = 1e15;
		while (l < r)
		
			long long mid = l + r + 1 >> 1;
			if (check(mid)) r = mid - 1;
			else l = mid;
		
		if (check(r)) cout << r << endl;
		else cout << r + 1 << endl;
	
	return 0;