AtCoder Beginner Contest 237（A-D）

A - Not Overflow

long long范围的数判断是否在int范围内，就可以强制转换

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 
using namespace std;
#define lowbit(x) ((x) & -(x))
#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;
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; }

const int INF = 0x3f3f3f3f;
const int mod = 1000000007;

void solve() {
	long long n;
	cin >> n;
	int m;
	m = n;
	if (m == n) {
		cout << "Yes" << endl;
	}
	else {
		cout << "No" << endl;
	}
}

int main() {
	IOS1;
	//IOS2;
	int __t = 1;
	//cin >> __t;
	for (int _t = 1; _t <= __t; _t++) {
		solve();
	}
	return 0;
}

B - Matrix Transposition

AC代码：

二维，数据大可能会被卡空间的

#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;
#define lowbit(x) ((x) & -(x))
#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;
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; }

const int INF = 0x3f3f3f3f;
const int mod = 1000000007;
void solve() {
	int h, w;
	cin >> h >> w;
	int x = max(h, w);
	vector> a(h + 5, vector(w + 5, 0));
	vector> b(w + 5, vector(h + 5, 0));
	for (int i = 1; i <= h; i++) {
		for (int j = 1; j <= w; j++) {
			cin >> a[i][j];
		}
	}
	for (int i = 1; i <= w; i++) {
		for (int j = 1; j <= h; j++) {
			b[i][j] = a[j][i];
		}
	}
	for (int i = 1; i <= w; i++) {
		for (int j = 1; j <= h; j++) {
			cout << b[i][j] << " n"[j == h];
		}
	}
}

int main() {
	IOS1;
	//IOS2;
	int __t = 1;
	//cin >> __t;
	for (int _t = 1; _t <= __t; _t++) {
		solve();
	}
	return 0;
}

一维，要看好各个行列关系

#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;
#define lowbit(x) ((x) & -(x))
#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;
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 T
read() {
	T sum = 0, fl = 1;
	int ch = getchar();
	for (; !isdigit(ch); ch = getchar())
		if (ch == '-') fl = -1;
	for (; isdigit(ch); ch = getchar()) sum = sum * 10 + ch - '0';
	return sum * fl;
}

const int INF = 0x3f3f3f3f;
const int mod = 1000000007;

void solve() {
	int h, w;
	cin >> h >> w;
	vector a(100010);
	for (int i = 1; i <= h * w; i++) {
		cin >> a[i];
	}
	for (int i = 1; i <= w; i++) {
		for (int j = 1; j <= h; j++) {
			cout << a[(j - 1) * w + i] << " n"[j == h];
		}
	}

}

int main() {
	IOS1;
	//IOS2;
	int __t = 1;
	//cin >> __t;
	for (int _t = 1; _t <= __t; _t++) {
		solve();
	}
	return 0;
}

C - kasaka

从开头添加若干个或零个a是否能成为回文字符串

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;
#define lowbit(x) ((x) & -(x))
#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;
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; }

const int INF = 0x3f3f3f3f;
const int mod = 1000000007;

void solve() {
	string s;
	cin >> s;
	int len = s.length();
	bool ok = true;
	for (int i = 0; i < len / 2; i++) {
		if (s[i] != s[len - 1 - i]) {
			ok = false;
		}
	}
	if (ok) {
		cout << "Yes" << endl;
	}
	else {
		int cnt = 0;
		for (int i = len - 1; i >= 0; i--) {
			if (s[i] == 'a') {
				cnt++;
			}
			else {
				break;
			}
		}
		int cnt1 = 0;
		for (int i = 0; i <= len - 1; i++) {
			if (s[i] == 'a') {
				cnt1++;
			}
			else {
				break;
			}
		}
		int cnt2 = abs(cnt1 - cnt);
		if (cnt1 >= cnt) {
			cout << "No" << endl;
			return;
		}
		string s1 = "";
		for (int i = 1; i <= cnt2; i++) {
			s1 += 'a';
		}
		s1 += s;
		ok = true;
		int len1 = s1.length();
		for (int i = 0; i < len1 / 2; i++) {
			if (s1[i] != s1[len1 - 1 - i]) {
				ok = false;
			}
		}
		if (ok) {
			cout << "Yes" << endl;
		}
		else {
			cout << "No" << endl;
		}
	}
}

int main() {
	IOS1;
	//IOS2;
	int __t = 1;
	//cin >> __t;
	for (int _t = 1; _t <= __t; _t++) {
		solve();
	}
	return 0;
}

D - LR insertion

根据题意就是往当前数的左边或者右边添值，乍一看从正面好像不太好弄，那就可以考虑逆着来，数组开个1e6不会太大，直接逆序添加就好了，L就放右边，R就放左边

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;
#define lowbit(x) ((x) & -(x))
#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;
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 T
read() {
	T sum = 0, fl = 1;
	int ch = getchar();
	for (; !isdigit(ch); ch = getchar())
		if (ch == '-') fl = -1;
	for (; isdigit(ch); ch = getchar()) sum = sum * 10 + ch - '0';
	return sum * fl;
}

const int INF = 0x3f3f3f3f;
const int mod = 1000000007;

void solve() {
	int n;
	cin >> n;
	vector a(1000010);
	string s;
	cin >> s;
	int len = s.size();
	int left = 500000 - 1, right = 500000 + 1;
	int x = len - 1;
	a[500000] = len;
	for (int i = len - 1; i >= 0; i--) {
		if (s[i] == 'L') {
			a[right++] = x;
			x--;
		}
		else {
			a[left--] = x;
			x--;
		}
	}
	for (int i = left + 1; i < right; i++) {
		cout << a[i] << " n"[i == right - 1];
	}
}

int main() {
	IOS1;
	//IOS2;
	int __t = 1;
	//cin >> __t;
	for (int _t = 1; _t <= __t; _t++) {
		solve();
	}
	return 0;
}

AtCoder Beginner Contest 237（A-D）

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