EatGrapes

There are multiple test cases. The first line of input contains an integer T (1 ≤ T ≤ 10), indicating the number of test cases. For each test case:

The first line contains an integer N (1 ≤ N ≤ $10^5$ ), indicating the number of nodes.

The second line contains N integers, $a_1$ , $a_2$ ,……, $a_n$ (0 ≤ $a_i$ ≤ $10^9$ ), indicating the number of grapes connected to the i-th node.

C(clang11) 解法, 执行用时: 118ms, 内存消耗: 4892K, 提交时间: 2021-01-15 13:36:54

#include<stdio.h>
#include<string.h>
#define maxn 102020
int a[maxn];
int main(){
	int n,m;
	int t;
	int sum;
	scanf("%d",&t);
	while(t--){
		sum=1;
		scanf("%d",&n);
		for(int i=1;i<=n;i++){
			scanf("%d",&a[i]);
		}
		a[n]++;
		for(int i=n;i>=1;i--){
			if(a[i]>1)break;
			else if(i!=1)sum=1-sum;
		}
		if(sum!=0)printf("these are sweet grapes\n");
		else printf("these are sour grapes\n");
	}
	return 0;
}

C++14(g++5.4) 解法, 执行用时: 65ms, 内存消耗: 372K, 提交时间: 2020-09-27 20:50:03

#include<stdio.h>
int main(){
	int T;
	scanf("%d", &T);
	while(T--){
		int flag = 0, n , x;
		scanf("%d",&n);
		for(int i = 1; i <= n; i++){
			scanf("%d", &x);
			if(x == 0)
				flag = 1 - flag;
			else
				flag = 1;
		}
		if(flag)
			printf("these are sweet grapes\n");
		else
			printf("these are sour grapes\n");
	}
	return 0;
}

C++11(clang++ 3.9) 解法, 执行用时: 146ms, 内存消耗: 388K, 提交时间: 2020-09-26 16:15:10

#include<bits/stdc++.h>
using namespace std;
int ans,T,n,x;
int main(){
	cin>>T;
	while(T--){
		cin>>n>>x;
		if(x==0)ans=0;
		else ans=1;
		for(int i=1;i<n;i++){
			cin>>x;
			if(x>0)ans=1;
			else ans=1-ans;
		}
		if(ans) cout<<"these are sweet grapes\n";
		else cout<<"these are sour grapes\n";
	}
	return 0;
}

详情