codeforce round＃784(div4) A-H

I just called cf A few days , If the explanation is not clear, please understand or leave a message in the comment area , And thank you for div4 The number of sessions makes me feel AK Happiness

Problem - A - CodeforcesCodeforces. Programming competitions and contests, programming communityhttps://codeforces.com/contest/1669/problem/AA There is no need to say more about the problem , Namely for+if else

inline void solve() {
	int n;
	scanf("%d", &n);
	if (n <= 1399) {
		printf("Division 4\n");
	} else if (n > 1399 && n <= 1599) {
		printf("Division 3\n");
	} else if (n > 1599 && n <= 1899) {
		printf("Division 2\n");
	} else {
		printf("Division 1\n");
	}
}

Problem - B - Codeforceshttps://codeforces.com/contest/1669/problem/B This problem is to count the numbers in the array , As long as a number appears 3 Once, you can directly output , It can be used map Realization

void solve() {
	int n;
	scanf("%d", &n);
	map<int, int> mp;
	int a;
	bool flag = true;
	for (int i = 1; i <= n; i++) {
		scanf("%d", &a);
		if (flag)
			if (++mp[a] >= 3) {
				printf("%d\n", a);
				flag = false;
				//break;
			}
	}
	if (flag)
		printf("-1\n");
}

Problem - C - CodeforcesCodeforces. Programming competitions and contests, programming communityhttps://codeforces.com/contest/1669/problem/C  Directly record the first two digits of the array , Odd digit and odd digit last1 Parity is different or even digits are different from last If the parity is different, you can never get a fully even or fully odd sequence

inline void solve() {
	int n;
	scanf("%d", &n);
	if (n == 2) {
		printf("YES\n");
		return;
	}
	int last1;
	int last2;
	scanf("%d %d", &last1, &last2);
	int t;
	bool flag = true;
	for (int i = 3; i <= n; i++) {
		scanf("%d", &t);
		if (flag)
			if (i & 1) {
				if (last1 % 2 != t % 2) {
					flag = false;
				}
			} else {
				if (last2 % 2 != t % 2) {
					flag = false;
				}
			}
	}
	if (flag) {
		printf("YES\n");
	} else {
		printf("NO\n");
	}
}

Problem - D - Codeforceshttps://codeforces.com/contest/1669/problem/D Can think of meeting in W The length of any previous sequence is greater than or equal to 2 Of BR Sequences can be mimeographed from lottery tickets , But only R or B The sequence cannot be printed , Then simulate this operation .

inline void solve() {
	int n;
	scanf("%d", &n);
	string s;
	cin >> s;
	if (n == 1 && s[0] != 'W') {
		printf("NO\n");
		return ;
	}
	int cnt = 0;
	char last;
	bool flag = true;
	map<char, int> mp;
	for (int i = 0; i < s.length(); i++) {
		mp[s[i]] = 1;
		if (flag)
			if (s[i] == 'W') {
				if (cnt == 1) {
					flag = false;
				}
				if ((mp['B'] + mp['R']) % 2) {
					flag = false;
				}
				cnt = 0;
				mp['B'] = 0;
				mp['R'] = 0;
			} else {
				cnt++;
			}
	}
	if (cnt == 1 || (mp['B'] + mp['R']) % 2) {
		flag = false;
	}
	if (flag) {
		printf("YES\n");
	} else {
		printf("NO\n");
	}
}
 

Problem - F - Codeforceshttps://codeforces.com/contest/1669/problem/F The first A The candy you can eat from the left map Prepare for it , Then start eating one by one from the right , If the number of sweets eaten by two people can be less than or equal to n And if they eat the same food together, update the answer , If two people eat the same amount, but the total number of two people exceeds n You can exit directly , Because it can't be established after .

ll s[MAXN];
inline void solve() {
	int n;
	scanf("%d", &n);
	map<ll, int> mp;
	ll sum = 0;
	for (int i = 1; i <= n; i++) {
		scanf("%lld", s + i);
		sum += s[i];
		mp[sum] = i;
	}
	int t;
	ll cnt = 0;
	bool flag = true;
	sum = 0;
	for (int i = n; i >= 1; i--) {
		sum += s[i];
		//printf("%d ", t);
		if (mp[sum]) {
			if (mp[sum] + n - i + 1 > n) {
				break;
			}
			cnt = mp[sum] + n - i + 1;
		}
	}
	printf("%lld\n", cnt);
}

Problem - E - Codeforceshttps://codeforces.com/contest/1669/problem/E Can pass map To record the number of first and second characters , It also records the number of times the string appears , When it is recorded to the i If a string is preceded by the same string as the first bit or the same string as the second bit, add , If this string has appeared before, subtract its *2, Because the first and second places are counted .

inline void solve() {
	int n;
	scanf("%d", &n);
	string s;
	map<string, int> mp;
	map<char, int> mp1;
	map<char, int> mp2;
	ll ans = 0;
	for (int i = 1; i <= n; i++) {
		cin >> s;
		ans += mp1[s[0]]++;
		ans += mp2[s[1]]++;
		ans -= mp[s] * 2;
		mp[s]++;
	}
	printf("%lld\n", ans);
}

Problem - H - CodeforcesCodeforces. Programming competitions and contests, programming communityhttps://codeforces.com/contest/1669/problem/H The meaning of the question is very clear, which is to start from the highest position and find out whether to change that position 1, First record the number of occurrences of each bit of each number in the array , Then start from the highest level to see whether the remaining operations can support its change 1

ll a[MAXN];
inline void solve() {
	int n, k;
	int b[50] = {0};
	scanf("%d %d", &n, &k);
	for (int i = 1; i <= n; i++) {
		scanf("%d", a + i);
		for (int j = 0; j <= 30; j++) {
			if ((a[i] >> j) & 1) {
				b[j]++;
			}
		}
	}
	int ans = 0;
	for (int i = 30; i >= 0; i--) {
		if (k >= n - b[i]) {
			k -= n - b[i];
			ans |= 1 << i;
		}
	}
	printf("%d\n", ans);
}