CodeForces 1102A

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本文解析了CodeForces1102A题目，介绍了一种高效算法，通过判断从1到n的整数序列之和的奇偶性来快速求解两集合元素差的最小绝对值，避免了超时问题。

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CodeForces 1102A
Integer Sequence Dividing

PROGRAM:
You are given an integer sequence 1,2,…,n1,2,…,n

. You have to divide it into two sets AA

and BB

in such a way that each element belongs to exactly one set and |sum(A)−sum(B)||sum(A)−sum(B)|

is minimum possible.The value |x||x|

is the absolute value of xx

and sum(S)sum(S)

is the sum of elements of the set SS

.InputThe first line of the input contains one integer nn

(1≤n≤2⋅1091≤n≤2⋅109

).OutputPrint one integer — the minimum possible value of |sum(A)−sum(B)||sum(A)−sum(B)|

if you divide the initial sequence 1,2,…,n1,2,…,n

into two sets AA

and BB

.ExamplesInputCopy3
OutputCopy0
InputCopy5
OutputCopy1
InputCopy6
OutputCopy1
NoteSome (not all) possible answers to examples:In the first example you can divide the initial sequence into sets A={1,2}A={1,2}

and B={3}B={3}

so the answer is 00

.In the second example you can divide the initial sequence into sets A={1,3,4}A={1,3,4}

and B={2,5}B={2,5}

so the answer is 11

.In the third example you can divide the initial sequence into sets A={1,4,5}A={1,4,5}

and B={2,3,6}B={2,3,6}

so the answer is 11

这道题题意是指输入一个n，把从1到n的所有数随意分成任意个数的两部分，使得这两部分之差最小。输出这个差值。

这样可以先从一些数据中找些规律。
当1~n的所有数据之和为偶数的时候，总能找到两对集合之和相等
而当所有数据之和为奇数的时候，则能找到两对集合之和之差为一。所以代码如下
但需要注意的是，如果求和用for循环的话，会超时。所以在求和是就能用等差数列求和公式进行运算。

#include <stdio.h>
int main()
{
	long long n,sum;
	while(~scanf("%d",&n))
	{
	sum=n*(n+1)/2;
	if(sum%2==0)
	printf("0\n");
	if(sum%2==1)
	printf("1\n");
 } 
}