codeforces-Increasing Sequence

本文描述了使用C++编程解决一个问题:给定一个数组,判断其元素是否可以通过排序构成严格递增序列,避免重复元素。作者通过先排序后遍历的方法实现判断。

题目意思是,给你一个数组,看数组中的元素,是否可以通过排序构成严格递增序列

简而言之,就是,给你的数组中会不会有重复的元素。

我的思路是,先用sort()排序,之后遍历数组,如果有相同的元素,就跳出循环,并输出"NO";

#include<cstdio>
#include"algorithm"
using namespace std;
int t,n,a[1100000];
int main()
{
	scanf("%d",&t);
	while(t--)
	{
		int flag=0;
		scanf("%d",&n);
		for(int i=0;i<n;i++)
			scanf("%d",&a[i]);
		sort(a,a+n);
		for(int i=0;i<n-1;i++)//这里需要注意,i的结束条件是<n-1;
		{
			if(a[i]==a[i+1])
			{
				flag=1;
				break;
			}	
		}
		if(flag==0)
			printf("YES\n");
		else
			printf("NO\n");	
	}
	return 0;
}

我举个栗子,就是数组a[3]={1,2,3};

它是a[0]先与a[1]比,之后a[1]再与a[2]比,之后没了(下标是从零开始的),我想表达的意思是数组的最后一个元素,是与前一个元素比较的,所以也就不可能出现a[2]与a[3]比较,也就没有i=n-1这一说

### Codeforces Problem 976C Solution in Python For solving problem 976C on Codeforces using Python, efficiency becomes a critical factor due to strict time limits aimed at distinguishing between efficient and less efficient solutions[^1]. Given these constraints, it is advisable to focus on optimizing algorithms and choosing appropriate data structures. The provided code snippet offers insight into handling string manipulation problems efficiently by customizing comparison logic for sorting elements based on specific criteria[^2]. However, for addressing problem 976C specifically, which involves determining the winner ('A' or 'B') based on frequency counts within given inputs, one can adapt similar principles of optimization but tailored towards counting occurrences directly as shown below: ```python from collections import Counter def determine_winner(): for _ in range(int(input())): count_map = Counter(input().strip()) result = "A" if count_map['A'] > count_map['B'] else "B" print(result) determine_winner() ``` This approach leverages `Counter` from Python’s built-in `collections` module to quickly tally up instances of 'A' versus 'B'. By iterating over multiple test cases through a loop defined by user input, this method ensures that comparisons are made accurately while maintaining performance standards required under tight computational resources[^3]. To further enhance execution speed when working with Python, consider submitting codes via platforms like PyPy instead of traditional interpreters whenever possible since they offer better runtime efficiencies especially important during competitive programming contests where milliseconds matter significantly.
评论 1
成就一亿技术人!
拼手气红包6.0元
还能输入1000个字符
 
红包 添加红包
表情包 插入表情
 条评论被折叠 查看
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值