AcWing每日一题

1801 蹄子剪刀布


分析：

• 暴力枚举六种情况，取最大值，比较容易想到

#include<bits/stdc++.h>
using namespace std;
const int N = 101;
int a[N];
int b[N];
//枚举
int main() {
	int n;
	cin>>n;
	for (int i = 0; i < n; i++) {
		cin >> a[i] >> b[i];
	}
	//枚举六种情况
	//1石头 2剪刀 3布
	int ans = 0;
	for (int i = 0; i < n; i++) {
		int m = a[i];
		int n = b[i];
		if ((m == 1 && n == 2) || (m == 2 && n == 3) || (m == 3 && n == 1)) {
			ans++;
		}
	}
	int ans1 = 0;
	//1石头 2布 3剪刀
	for (int i = 0; i < n; i++) {
		int m = a[i];
		int n = b[i];
		if ((m == 1 && n == 3) || (m == 2 && n == 1) || (m == 3 && n == 2)) {
			ans1++;
		}
	}
	int win = max(ans, ans1);
	//1剪刀 2布 3石头
	int ans2 = 0;
	for (int i = 0; i < n; i++) {
		int m = a[i];
		int n = b[i];
		if ((m == 1 && n == 2) || (m == 2 && n == 3) || (m == 3 && n == 1)) {
			ans2++;
		}
	}
	win = max(win, ans2);
	//1剪刀 2石头  3布
	int ans3 = 0;
	for (int i = 0; i < n; i++) {
		int m = a[i];
		int n = b[i];
		if ((m == 1 && n == 3) || (m == 2 && n == 1) || (m == 3 && n == 2)) {
			ans3++;
		}
	}
	win = max(win, ans3);
	//1布 2剪刀 3石头
	int ans4 = 0;
	for (int i = 0; i < n; i++) {
		int m = a[i];
		int n = b[i];
		if ((m == 1 && n == 3) || (m == 2 && n == 1) || (m == 3 && n == 2)) {
			ans4++;
		}
	}
	win = max(win, ans4);
	int ans5 = 0;
	//1布 2石头 3剪刀 
	for (int i = 0; i < n; i++) {
		int m = a[i];
		int n = b[i];
		if ((m == 1 && n == 2) || (m == 2 && n == 3) || (m == 3 && n == 1)) {
			ans5++;
		}
	}
	win=max(win,ans5);
     cout<<win<<endl;
}

• 看了别人写的题解，第二种思路很巧妙，我们发现
  -无论1，2，3代表那种手势， (1,2) (2,3) (1,3)之间都存在胜负关系，当我们把编号都减去 1 后，也即三种手势分别为 (0,1,2)时，可以通过 (x+1)%3 获取 x手势的下一个相邻手势
  比如(0+1)%3=1 (1+1)%3=2 (2+1)%3=0
  然后用a b去记录两种胜负情况，取最大值即可

#include <bits/stdc++.h>
using namespace std;
int main() {

    int n;
    cin >> n;
    int a = 0, b = 0;
    for(int i = 0; i < n; ++i) {
        int x, y;
        cin >> x >> y;
        x--, y--;
        // 当 x + 1 能赢 x 时
        if((x + 1) % 3 == y) a++;
        // 当 x 能赢 x + 1 时
        if(x == (y + 1) % 3) b++;
    } 
    cout << max(a, b) << endl;
    return 0;
}

作者：算法小爱
链接：https://www.acwing.com/solution/content/88041/
来源：AcWing
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