CodeForces - 849

本文解析并实现了三道算法题目,包括简单的数组判断、寻找等斜率直线以及字符串价值计算,涉及C语言基础、数组操作及数学计算。

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A. Odds and Ends

简单判断

#include<cstdio>
int a[105];

int main(){
	int n;
	while(~scanf("%d",&n)){
		int cnt=0;
		for(int i=0;i<n;i++)
			scanf("%d",&a[i]);
		
		
		if(n%2==0||a[0]%2==0||a[n-1]%2==0)
			printf("No\n");
				
		if(n%2&&a[0]%2&&a[n-1]%2)
			printf("Yes\n");
			
	}
	return 0;
} 

B. Tell Your World

斜率k是点1,2,3构成三种直线的其一斜率,然后寻找两条等斜率的基准点逐点判断

#include<cstdio>
const int N=1005;
int p[N];
int n;

bool work(double k){
	int flag=0,next=-1;
	for(int i=2;i<=n;i++){
		if(p[i]-p[1]==k*(i-1)) continue; //1,i
		flag=1;  //other
		if(next<0) next=i;  
		else if(p[i]-p[next]!=k*(i-next)){
			flag=0;
			break;
		}
	}
	if(flag) return true;
	else return false;
}

int main(){
	while(~scanf("%d",&n)){
		for(int i=1;i<=n;i++)
			scanf("%d",&p[i]);
		
		double k1=1.0*(p[2]-p[1]);  //1,2
		double k2=0.5*(p[3]-p[1]);  //1,3
		double k3=1.0*(p[3]-p[2]);  //2,3
		
		if(work(k1)||work(k2)||work(k3))
			printf("Yes\n");
		else
			printf("No\n");
	}
	return 0;
}

C. From Y to Y

一个字符串中有n个a的话,那么对应这些字符a贡献的价值是:(n*(n-1))/2

#include<cstdio>
#include<cstring>
const int N=100;

int main(){
	int n;
	while(~scanf("%d",&n)){
		int cost=0;
		for(int i=1;i<=26;i++){
			int pos=100;
			for(int j=1;j<=N;j++){
				if(cost+j*(j-1)/2>n){
					pos=j-1;
					break;
				}
			}
			
			for(int k=1;k<=pos;k++)
				printf("%c",i+'a'-1);
			cost+=(pos-1)*pos/2;
			if(cost==n) break;
		}
		printf("\n");
	}
	return 0;
}

 

### 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.
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