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");
}
}