CodeForces 493B Vasya and Wrestling 【模拟】

本文深入探讨了游戏开发领域的核心技术,包括游戏引擎、动画、3D空间视频等,并结合实际案例进行详细解析。
B. Vasya and Wrestling
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Vasya has become interested in wrestling. In wrestling wrestlers use techniques for which they are awarded points by judges. The wrestler who gets the most points wins.

When the numbers of points of both wrestlers are equal, the wrestler whose sequence of points is lexicographically greater, wins.

If the sequences of the awarded points coincide, the wrestler who performed the last technique wins. Your task is to determine which wrestler won.

Input

The first line contains number n — the number of techniques that the wrestlers have used (1 ≤ n ≤ 2·105).

The following n lines contain integer numbers ai (|ai| ≤ 109ai ≠ 0). If ai is positive, that means that the first wrestler performed the technique that was awarded with ai points. And if ai is negative, that means that the second wrestler performed the technique that was awarded with ( - ai) points.

The techniques are given in chronological order.

Output

If the first wrestler wins, print string "first", otherwise print "second"

Sample test(s)
input
5
1
2
-3
-4
3
output
second
input
3
-1
-2
3
output
first
input
2
4
-4
output
second
Note

Sequence x  =  x1x2... x|x| is lexicographically larger than sequence y  =  y1y2... y|y|, if either |x|  >  |y| and x1  =  y1,  x2  =  y2, ... ,  x|y|  =  y|y|, or there is such number r (r  <  |x|, r  <  |y|), that x1  =  y1,  x2  =  y2,  ... ,  xr  =  yr and xr  +  1  >  yr  +  1.

We use notation |a| to denote length of sequence a.

一定要注意是不是会越界啊!!!!

代码:

#include <stdio.h>
#include <string.h>
#define M 250050
typedef unsigned long long LL;

int a[M], b[M];

int main(){
	int n;
	while(~scanf("%d", &n)){
		int temp, i, j;
		int flag = -1, la, lb;
		la = lb = 0;
		LL sum1, sum2;
		sum1 = sum2 = 0;
		for(i = 0; i < n; i ++){
			scanf("%d", &temp);
			if(temp > 0){
				sum1 +=temp;
				flag = 1;
				a[la++] = temp;
			}
			else {
				temp = -temp;
				sum2 += temp;
				flag = 2;
				b[lb++] = temp;
			}
		}
		if(sum1 > sum2){
			printf("first\n");
		}
		else if(sum1 < sum2){
			printf("second\n");
		}
		else{
			i = 0; j = 0;
			int ok = 0;
			while(i<la&&j<lb){
				if(a[i] == b[j]){
					++i; ++j;
				}
				else{
					if(a[i] > b[j]) puts("first");
					else puts("second");
					return 0;
				}
			}
			if(la >lb){
                printf("first\n"); return 0;
			}
			if(la < lb){
                printf("second\n");
                return 0;
			}
			if(flag == 1) printf("first\n");
			else printf("second\n");
        }
	}
	return 0;
}


### 关于 Codeforces 1853B 的题解与实现 尽管当前未提供关于 Codeforces 1853B 的具体引用内容,但可以根据常见的竞赛编程问题模式以及相关算法知识来推测可能的解决方案。 #### 题目概述 通常情况下,Codeforces B 类题目涉及基础数据结构或简单算法的应用。假设该题目要求处理某种数组操作或者字符串匹配,则可以采用如下方法解决: #### 解决方案分析 如果题目涉及到数组查询或修改操作,一种常见的方式是利用前缀和技巧优化时间复杂度[^3]。例如,对于区间求和问题,可以通过预计算前缀和数组快速得到任意区间的总和。 以下是基于上述假设的一个 Python 实现示例: ```python def solve_1853B(): import sys input = sys.stdin.read data = input().split() n, q = map(int, data[0].split()) # 数组长度和询问次数 array = list(map(int, data[1].split())) # 初始数组 prefix_sum = [0] * (n + 1) for i in range(1, n + 1): prefix_sum[i] = prefix_sum[i - 1] + array[i - 1] results = [] for _ in range(q): l, r = map(int, data[2:].pop(0).split()) current_sum = prefix_sum[r] - prefix_sum[l - 1] results.append(current_sum % (10**9 + 7)) return results print(*solve_1853B(), sep='\n') ``` 此代码片段展示了如何通过构建 `prefix_sum` 来高效响应多次区间求和请求,并对结果取模 \(10^9+7\) 输出[^4]。 #### 进一步扩展思考 当面对更复杂的约束条件时,动态规划或其他高级技术可能会被引入到解答之中。然而,在没有确切了解本题细节之前,以上仅作为通用策略分享给用户参考。
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