2024“钉耙编程”中国大学生算法设计超级联赛（9）

传送门

1005 怪物猎人

如果全用 x 和 全用 y 攻击的轮次不同那么奇偶轮次都可以击败怪物；否则只有奇数轮次或者偶数轮次能够击败怪物。

#include <bits/stdc++.h>
using namespace std;

#define int long long

inline int read() {
	int x = 0, f = 1; char c = getchar();
	while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar(); }
	while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return x * f;
}

signed main() {
	int t = read();
	while (t--) {
		int k = read(), x = read(), y = read();
		int r1 = (k + x - 1) / x, r2 = (k + y - 1) / y;
		if (r1 != r2) printf("Yes\nYes\n");
		else if (r1 & 1) printf("Yes\nNo\n");
		else printf("No\nYes\n");
	}
	return 0;
}

1007 小猫钓鱼

如果两个人手里拿着的都是完整的对子，那么是 sha7dow 赢；否则就是 shuishui 赢。

#include <bits/stdc++.h>
using namespace std;

const int N = 2e5 + 10;
int a[N], b[N], vis[N];

inline int read() {
	int x = 0, f = 1; char c = getchar();
	while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar(); }
	while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
	return x * f;
}

int main() {
	int t = read();
	while (t--) {
		int n = read(), flag = 0;
		for (int i = 1; i <= n; i++) vis[i] = 0;
		for (int i = 1; i <= n; i++) {
			a[i] = read();
			if (vis[a[i]]) flag++;
			vis[a[i]] = 1;
		}
		for (int i = 1; i <= n; i++) {
			b[i] = read();
		}
		if (!flag) puts("sha7dow");
		else puts("shuishui");
	}
	return 0;
}