2019年广东工业大学腾讯杯新生程序设计竞赛

题目链接：https://ac.nowcoder.com/acm/contest/3036/
题解链接：https://www.nowcoder.com/discuss/352755

A - 原初的信纸

#include <iostream>
using namespace std;

int main(void)
{
	int t, n, minv, val;
	cin >> t;
	while (t--){
		minv = 0;
		cin >> n;
		while (n--){
			cin >> val;
			minv = max(minv, val);
		}
		cout << minv << endl;
	}
	return 0;
}

B - 骑士的对决

#include <iostream>
#include <string>
using namespace std;
//S J B
int main(void)
{
	char a, b, c;
	cin >> a >> b >> c;
	
	if (c == 'S'){
		if (a == 'B' || b == 'B')
			cout << "pmznb" << endl;
		else
			cout << "lyrnb" << endl;
	}
	else if (c == 'J'){
		if (a == 'S' || b == 'S')
			cout << "pmznb" << endl;
		else
			cout << "lyrnb" << endl;
	}
	else if (c == 'B'){
		if (a == 'J' || b == 'J')
			cout << "pmznb" << endl;
		else
			cout << "lyrnb" << endl;
	}
	
	return 0;
}

C - 秘密的议会

#include <iostream>
#include <string>
using namespace std;

int main(void)
{
	int t, y, n;
	string s;
	cin >> t;
	while (t--){
		y = n = 0;
		cin >> s;
		for (int i = 0; i < s.size(); i++){
			if (s[i] == 'y' || s[i] == 'Y')
				y++;
			else if (s[i] == 'n' || s[i] == 'N')
				n++;
		}
		if (y >= s.size() / 2)
			cout << "pmznb" << endl;
		else if (n >= s.size() / 2)
			cout << "lyrnb" << endl;
		else
			cout << "wsdd" << endl;
	}
	return 0;
}

D - 城市的税金

在进行区间询问的时候，因为数的范围为1e9，不能用数组来记录每个数出现的次数
但是可以将区间中的数字放到另一个数组中，进行排序，找出最多有几个数连续即可。

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
typedef long long ll;

ll a[105], b[105];

void change(int idx)
{
	a[idx] = a[idx] * 251 % 996 * 404 * 123;
}

int main(void)
{
	int n, m, t, l, r;
	cin >> n >> m;
	for (int i = 1; i <= n; i++)
		cin >> a[i];
	while (m--){
		cin >> t;
		if (t == 1){
			cin >> l >> r;
			for (int i = l; i <= r; i++)
				change(i);
		}
		else{
			cin >> l >> r;
			for (int i = l; i <= r; i++){
				b[i] = a[i];
			}
			sort(b + l, b + r + 1);
			int maxv = 1, cnt = 1;
			for (int i = l + 1; i <= r; i++){
				if (b[i] == b[i - 1])
					cnt++;
				else
					cnt = 1;
				maxv = max(maxv, cnt);
			}
			cout << maxv << endl;
		}
	}
	
	return 0;
}

K - 消亡的质数

$因式分解步骤：$
$x^3-y^3$
$=(x^3-x^2y)+x^2y-(y^3-xy^2)-xy^2$
$=x^2(x-y)-y^2(y-x)+xy(x-y)$
$=(x-y)(x^2+xy+y^2)$

每次做数论题都感觉好神奇，数论只会gcd QAQ

M - 破碎的愿望

注意特判能被2n整除的情况

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;

char str[100];

int main(void)
{
	ll n, k;
	cin >> n >> k;
	cin >> str + 1;
	for (int i = 1; i <= n; i++)
		str[2 * n - i + 1] = str[i];
	if (k % (2 * n) == 0)
		cout << str[1] << endl;
	else
		cout << str[k % (2 * n)] << endl;
	
	return 0;
}