Educational Codeforces Round 115 (Rated for Div. 2)题解

好长时间没有写过题解了，以后还是坚持写吧！！！

A

题意：给一个2*n的矩阵，起点是(1, 1)，终点是(2, n)，矩阵中的数是由0和1组成的move from cell (x1,y1) to cell (x2,y2) in one step if |x1−x2|≤1 and |y1−y2|≤1，可以左右、上下也可以斜着走，1表示不可以通过，0表示可以过，问可不可以到达终点。
思路：找上下两个数如果都是1的话，就表示不能通过。
代码

#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
int t, n;
int main()
{
	cin >> t;
	while (t -- ){
		cin >> n;
		string a1, a2;
		cin >> a1 >> a2;
		int flag = 0;
		for (int i = 0; i < n; i ++ ) {
			if (a1[i] == a2[i] && a1[i] == '1') {
				cout << "NO" << endl;
				flag = 1;
				break;
			}
		}
		if (flag == 0) cout <<"YES" << endl ;
	}
	return 0;
}

B

题意：有$n$个人（偶数），这$n$个人要分成两组，每组的人数一样多是$n/2$，每周每组必须上一节课，而且两组上课不是同一天，给出每个人每天是否空闲，看是否可以分出来两组。
思路：暴力思路，看每个天是否大于等于$n/2$个人，然后看可不可以分成两组。
代码：

#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
int t, n, a[N][10], st[N], backup[N];
int main()
{
	cin >> t;
	while (t -- ) {
		cin >> n;
		for (int i = 1; i <= n; i ++ ) 
			for (int j = 1; j <= 5; j ++ ) 
				cin >> a[i][j] ;
		vector<int>s[7];
		for (int j = 1; j <= 5; j ++ ) {
			for (int i = 1; i <= n; i ++ ) {
				if (a[i][j] == 1) s[j].push_back(i);
			}
		}
		int flag = 0;
		//for (int i = 1; i <= 5; i ++ ) cout << s[i].size() << " ";
		//cout << endl;
		for (int k = 1; k <= 5; k ++ ) {
			for (int i = 1; i <= n; i ++ ) st[i] = 0;
			if (s[k].size() >= n / 2) {
				for (int i = 0; i < s[k].size(); i ++ ) st[s[k][i]] = 1;
				for (int j = k + 1; j <= 5; j ++ ) {
					if (s[j].size() >= n / 2) {
						int u = 0;
						memcpy(backup, st, sizeof st);
						for (int x = 0; x < s[j].size(); x ++ ) st[s[j][x]] = 1;
						for (int x = 1; x <= n; x ++ ) {
							if (st[x] == 0) {
								u = 1;
								break;
							}
						}
						if (u == 0) {
							cout << "YES" << endl;
							flag = 1;
							break;
						}
						memcpy(st, backup, sizeof st);
					}
				}
			}
			if (flag == 1) break;
		}
		if (flag == 0) cout << "NO" << endl;
	}
	return 0;
}

C

题意：给一个长度为n的数组，平均值为k（k为实数），在这个数组里面去2个数，剩下的平均值还是k，问有多少对这样的数。
思路：暴力枚举二分
代码：

#include <bits/stdc++.h>
using namespace std;
const int N = 2e5+ 10;
long long t, n, sum;
int a[N], b[N];
int main()
{
	cin >> t;
	while (t -- ) {
		cin >> n;
		sum = 0;
		for (int i = 1; i <= n; i ++ ) {
			cin >> a[i];
			sum += a[i];
		}
		for (int i = 1; i <= n; i ++ ) b[i] = 0;
		b[n] = 1;
		sort(a + 1, a + n + 1) ;
		for (int i = n - 1; i >= 1; i -- ) {
			if (a[i] != a[i + 1]) {
				b[i] = 1;
			}
			else b[i] = b[i + 1] + 1;
		}
		//for (int i = 1; i <= n; i ++ ) cout << b[i] << " " ;
		//cout << endl ;
		int k = floor((double)sum * (double)2 / (double)n);
		long long num = 0 ;
		if (k != (double)sum * (double)2 / (double)n ) {
			cout << 0 << endl;
			continue;
		}
		for (int i = 1; i <= n; i ++ ) {
			int d = lower_bound(a + i + 1, a + n + 1, (long long)k - a[i]) - a;
			//t
			//while (a[i] + a[d] == k && d <= n) d ++;
			//cout << i << " " << d << " " << b[d] << endl;
			if (d != i && d <= n && a[d] + a[i] == k)num += b[d];
		}
		cout << num <<  endl ;
	}
	return 0;
}

D

题意：从n道题中选择3道题，每道题都有一个主题和一个难度，最少满足两个条件的其中一个，求所有的组合数。
思路：所有的组合去除不满足要求的组合。
代码：

#include <bits/stdc++.h>
using namespace std ;
const int N = 2e5 + 10 ;
int a[N][5] ;
int T, n ;
int st1[N], st2[N] ;
int main()
{
	cin >> T ;
	while (T -- ) {
		cin >> n ;
		for (int i = 1; i <= n; i ++ ) 
			st1[i] = 0, st2[i] = 0 ; 
		for (int i = 1 ; i <= n; i ++ ) { 
			cin >> a[i][1] >> a[i][2];
			st1[a[i][1]] ++ ;
			st2[a[i][2]] ++ ;
		}
		//所有的组合 C(n, 3) 
		long long ans = (long long) n * (n - 1) * (n - 2) / 6 ;
		//不满足要求的组合 
		for (int i = 1; i <= n; i ++ ) 
			ans -= (long long) (st1[a[i][1]] - 1) * (st2[a[i][2]] - 1) ;
		cout << ans << endl ;
	}
	return 0 ;
}