CF1102A Integer Sequence Dividing

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    • 130提交
  • 题目来源 CodeForces 1102A
  • 评测方式 RemoteJudge
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  • 难度 普及/提高-
  • 时空限制 1000ms / 256MB

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    题意翻译

    给你一个数n,求把1~n的整数分为两组,使得每组的和的差的绝对值最小。

    题目描述

    You are given an integer sequence $ 1, 2, \dots, n $ . You have to divide it into two sets $ A $ and $ B $ in such a way that each element belongs to exactly one set and $ |sum(A) - sum(B)| $ is minimum possible.

    The value $ |x| $ is the absolute value of $ x $ and $ sum(S) $ is the sum of elements of the set $ S $ .

    输入输出格式

    输入格式:

    The first line of the input contains one integer $ n $ ( $ 1 \le n \le 2 \cdot 10^9 $ ).

    输出格式:

    Print one integer — the minimum possible value of $ |sum(A) - sum(B)| $ if you divide the initial sequence $ 1, 2, \dots, n $ into two sets $ A $ and $ B $ .

    输入输出样例

    输入样例#1: 复制
    3
    
    输出样例#1: 复制
    0
    
    输入样例#2: 复制
    5
    
    输出样例#2: 复制
    1
    
    输入样例#3: 复制
    6
    
    输出样例#3: 复制
    1
    

    说明

    Some (not all) possible answers to examples:

    In the first example you can divide the initial sequence into sets $ A = \{1, 2\} $ and $ B = \{3\} $ so the answer is $ 0 $ .

    In the second example you can divide the initial sequence into sets $ A = \{1, 3, 4\} $ and $ B = \{2, 5\} $ so the answer is $ 1 $ .

    In the third example you can divide the initial sequence into sets $ A = \{1, 4, 5\} $ and $ B = \{2, 3, 6\} $ so the answer is $ 1 $ .

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