【CodeForces 282B】Painting Eggs(模拟)

本文介绍了一个关于如何公平分配鸡蛋的问题,并提供了一种通过编程模拟的方法来解决该问题。问题中涉及两个孩子A和G,他们对每个鸡蛋有不同的定价。文章提供了一个C++程序示例,用于找到一种分配方式使得两方获得的总金额之差不超过500。

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B. Painting Eggs
time limit per test
5 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

The Bitlandians are quite weird people. They have very peculiar customs.

As is customary, Uncle J. wants to have n eggs painted for Bitruz (an ancient Bitland festival). He has asked G. and A. to do the work.

The kids are excited because just as is customary, they're going to be paid for the job!

Overall uncle J. has got n eggs. G. named his price for painting each egg. Similarly, A. named his price for painting each egg. It turns out that for each egg the sum of the money both A. and G. want for the painting equals 1000.

Uncle J. wants to distribute the eggs between the children so as to give each egg to exactly one child. Also, Uncle J. wants the total money paid to A. to be different from the total money paid to G. by no more than 500.

Help Uncle J. Find the required distribution of eggs or otherwise say that distributing the eggs in the required manner is impossible.

Input

The first line contains integer n (1 ≤ n ≤ 106) — the number of eggs.

Next n lines contain two integers ai and gi each (0 ≤ ai, gi ≤ 1000; ai + gi = 1000): ai is the price said by A. for the i-th egg and gi is the price said by G. for the i-th egg.

Output

If it is impossible to assign the painting, print "-1" (without quotes).

Otherwise print a string, consisting of n letters "G" and "A". The i-th letter of this string should represent the child who will get the i-th egg in the required distribution. Letter "A" represents A. and letter "G" represents G. If we denote the money Uncle J. must pay A. for the painting as Sa, and the money Uncle J. must pay G. for the painting as Sg, then this inequality must hold: |Sa  -  Sg|  ≤  500.

If there are several solutions, you are allowed to print any of them.

Examples
Input
2
1 999
999 1
Output
AG
Input
3
400 600
400 600
400 600
Output
AGA

题目大意:有n个鸡蛋,分给A和G两个人,每个人的对每个鸡蛋的价格不同,如何分配能保证|Sa - Sg|<=500
思路:不太明白题目的用意是什么,两个和为1000有什么用,反正我就模拟了一下,可能有什么深意没领会到吧
#include<iostream>
#include<cstdio>     
#include<cstdlib>
#define manx 1000005
using namespace std;
int main()
{
    int n,suma=0,sumg=0,a,g;
    int cot=0;
    char s[manx];
    scanf("%d",&n);
    for (int i=0; i<n; i++){
        scanf("%d%d",&a,&g);
        if(suma + a > sumg + 500){
            sumg+=g;
            s[cot++]='G';
        }
        else{
            suma+=a;
            s[cot++]='A';
        }
    }
    if(abs(suma - sumg) > 500){
        printf("-1\n");
        return 0;
    }
    for (int i=0; i<cot; i++)
        printf("%c",s[i]);
    printf("\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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