Georgia and Bob

The first line of the input contains a single integer T $(1 \leq T \leq 20)$ , the number of test cases. Then T cases follow. Each test case contains two lines. The first line consists of one integer N $(1 \leq N \leq 1000)$ , indicating the number of chessmen. The second line contains N different integers $P_1, P_2 ... P_n$ $(1 \leq P_i \leq 10000)$ , which are the initial positions of the n chessmen.

C++14(g++5.4) 解法, 执行用时: 3ms, 内存消耗: 492K, 提交时间: 2019-11-02 22:06:19

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

int main()
{
	int t;
	cin >> t;
	while( t-- )
	{
		int a[1005];
		int n;
		cin >> n;
		for (int i = 0; i < n; i++)
		{
			cin >> a[i];
		}
		int ans = 0;
		sort(a,a+n);
		for (int i = n - 1; i >= 0; i-= 2)
		{
			if( i == 0 )
			{
				if( a[i] != 1 ) ans ^= (a[i] - 1);
			}
			else ans ^= (a[i] - a[i-1] - 1); 
		}
		if( ans == 0 ) cout << "Bob will win" << endl;
		else cout << "Georgia will win" << endl;
	}
	return 0;
}

C++ 解法, 执行用时: 3ms, 内存消耗: 284K, 提交时间: 2021-08-08 11:40:05

#include <iostream>
#include <algorithm>
using namespace std;
const int MAXN = 1e3 + 7;
int a[MAXN];
int main()
{
	int t;
	scanf("%d",&t);
	while(t--){
		int n;
		scanf("%d",&n);
		for(int i = 1;i <= n;++i)	scanf("%d",&a[i]);
		sort(a + 1,a + n + 1);
		int ans = 0;
		for(int i = n;i > 0;i -= 2)
		ans ^= (a[i] - a[i - 1] - 1);
		if(ans)
		puts("Georgia will win");
		else
		puts("Bob will win");
	}
	return 0;
}

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