Codeforces Round #773 (Div. 2)（A-C）（multiset应用）

Problem - A - Codeforces

当所有点的y坐标都不一样的时候，可以满足题意，当有两个点的y坐标相同时，如果另一个点y坐标在其上方则满足题意，否则不满足，输出两个点之间的距离

AC代码：

// #pragma GCC optimize(2)
// #pragma GCC optimize(3)
// #pragma GCC optimize("Ofast")
// #include
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
// using i64 = long long;
#define lowbit(x) ((x) & -(x))
#define endl 'n'
#define IOS1 ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define IOS2 ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
typedef vector vi;
typedef vector vll;
typedef vector vc;
typedef long long ll;
typedef long long i64;
template T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }
template T lcm(T a, T b) { return a / gcd(a, b) * b; }
template
T power(T a, int b) {
	T res = 1;
	for (; b; b >>= 1, a = a * a) {
		if (b & 1) {
			res = res * a;
		}
	}
	return res;
}
template 
inline void read(T& x)
{
	x = 0; int f = 1; char ch = getchar();
	while (!isdigit(ch)) { if (ch == '-') f = -1; ch = getchar(); }
	while (isdigit(ch)) { x = x * 10 + ch - '0', ch = getchar(); }
	x *= f;
}
const int INF = 0x3f3f3f3f;
const int mod = 1000000007;
const double PI = acos(-1.0);
const double eps = 1e-6;
inline int sgn(double x) {
	return x < -eps ? -1 : x > eps;
}

void solve() {
	vector> a(5);
	for (int i = 1; i <= 3; i++) {
		cin >> a[i].first >> a[i].second;
	}
	if (a[1].second == a[2].second) {
		if (a[3].second < a[1].second) {
			cout << max(a[1].first, a[2].first) - min(a[1].first, a[2].first) << endl;
		}
		else {
			cout << 0 << endl;
		}
	}
	else if (a[1].second == a[3].second) {
		if (a[2].second < a[1].second) {
			cout << max(a[1].first, a[3].first) - min(a[1].first, a[3].first) << endl;
		}
		else {
			cout << 0 << endl;
		}
	}
	else if (a[2].second == a[3].second) {
		if (a[1].second < a[2].second) {
			cout << max(a[2].first, a[3].first) - min(a[2].first, a[3].first) << endl;
		}
		else {
			cout << 0 << endl;
		}
	}
	else {
		cout << 0 << endl;
	}
	return;
}
int main() {
	IOS1;
	//IOS2;
	int __t = 1;
	cin >> __t;
	for (int _t = 1; _t <= __t; _t++) {
		solve();
	}
	return 0;
}

Problem - B - Codeforces

当某个数出现次数是1的时候，只要往外拿结果不变，只有当某个数出现次数大于1，当它往外拿的时候，对应的个数就会+1

AC代码：

// #pragma GCC optimize(2)
// #pragma GCC optimize(3)
// #pragma GCC optimize("Ofast")
// #include
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
// using i64 = long long;
#define lowbit(x) ((x) & -(x))
#define endl 'n'
#define IOS1 ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define IOS2 ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
typedef vector vi;
typedef vector vll;
typedef vector vc;
typedef long long ll;
typedef long long i64;
template T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }
template T lcm(T a, T b) { return a / gcd(a, b) * b; }
template
T power(T a, int b) {
	T res = 1;
	for (; b; b >>= 1, a = a * a) {
		if (b & 1) {
			res = res * a;
		}
	}
	return res;
}
template 
inline void read(T& x)
{
	x = 0; int f = 1; char ch = getchar();
	while (!isdigit(ch)) { if (ch == '-') f = -1; ch = getchar(); }
	while (isdigit(ch)) { x = x * 10 + ch - '0', ch = getchar(); }
	x *= f;
}
const int INF = 0x3f3f3f3f;
const int mod = 1000000007;
const double PI = acos(-1.0);
const double eps = 1e-6;
inline int sgn(double x) {
	return x < -eps ? -1 : x > eps;
}

void solve() {
	int n;
	cin >> n;
	map mp;
	for (int i = 0; i < n; i++) {
		int x;
		cin >> x;
		mp[x]++;
	}
	int cnt = mp.size();
	for (int i = 1; i <= n; i++) {
		cout << max(i, cnt) << " n"[i == n];
	}
	return;
}
int main() {
	IOS1;
	//IOS2;
	int __t = 1;
	cin >> __t;
	for (int _t = 1; _t <= __t; _t++) {
		solve();
	}
	return 0;
}

Problem - C - Codeforces

multiset应用，c++语言中，multiset是库中一个非常有用的类型，它可以看成一个序列，插入一个数，删除一个数都能够在O(logn)的时间内完成，而且他能时刻保证序列中的数是有序的，而且序列中可以存在重复的数。该题把题意读懂就好了，根据贪心，先对序列进行从小到大排序，这样前面的数*x后一定会得到后面的数，就不用再回来浪费时间了

AC代码：

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#include
// #include 
// #include 
// #include 
// #include 
// #include 
// #include 
// #include 
// #include 
// #include 
// #include 
// #include 
// #include 
// #include 
// #include 
// #include 
// #include 
using namespace std;
// using i64 = long long;
#define lowbit(x) ((x) & -(x))
#define endl 'n'
#define IOS1 ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define IOS2 ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
typedef vector vi;
typedef vector vll;
typedef vector vc;
typedef long long ll;
typedef long long i64;
template T gcd(T a, T b) { return b ? gcd(b, a % b) : a; }
template T lcm(T a, T b) { return a / gcd(a, b) * b; }
template
T power1(T a, int b) {
	T res = 1;
	for (; b; b >>= 1, a = a * a) {
		if (b & 1) {
			res = res * a;
		}
	}
	return res;
}
template
T power2(T a, int b, T c) {
	T res = 1;
	for (; b; b >>= 1, a = a * a % c) {
		if (b & 1) {
			res = res * a % c;
		}
	}
	return res;
}
template 
T Myabs(T a) {
	return a >= 0 ? a : -a;
}
template 
inline void read(T& x)
{
	x = 0; int f = 1; char ch = getchar();
	while (!isdigit(ch)) { if (ch == '-') f = -1; ch = getchar(); }
	while (isdigit(ch)) { x = x * 10 + ch - '0', ch = getchar(); }
	x *= f;
}
const int INF = 0x3f3f3f3f;
const int mod = 1000000007;
// const int mod = 998244353;
const double PI = acos(-1.0);
const double eps = 1e-6;
inline int sgn(double x) {
	return x < -eps ? -1 : x > eps;
}


void solve() {
	int n, x;
	cin >> n >> x;
	vector a(n);
	for (int i = 0; i < n; i++) {
		cin >> a[i];
	}
	multiset s;
	sort(a.begin(), a.end());
	for (auto v : a) {
		if (v % x == 0 && s.find(v / x) != s.end()) {
			s.erase(s.find(v / x));
		}
		else {
			s.insert(v);
		}
	}
	cout << s.size() << endl;
	return;
}
signed main() {
	IOS1;
	//IOS2;
	int __t = 1;
	cin >> __t;
	for (int _t = 1; _t <= __t; _t++) {
		solve();
	}
	return 0;
}

multiset:

Codeforces Round #773 (Div. 2)（A-C）（multiset应用）

C/C++/C#相关栏目本月热门文章