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本文链接：https://blog.csdn.net/weixin_42449444/article/details/100170399
Problem Description：
You are given n chips on a number line. The i-th chip is placed at the integer coordinate
. Some chips can have equal coordinates.
You can perform each of the two following types of moves any (possibly, zero) number of times on any chip:
- Move the chip i by 2 to the left or 2 to the right for free (i.e. replace the current coordinate
with
or with
);
- move the chip i by 1 to the left or 1 to the right and pay one coin for this move (i.e. replace the current coordinate
with
or with
).
Note that it's allowed to move chips to any integer coordinate, including negative and zero.
Your task is to find the minimum total number of coins required to move all n chips to the same coordinate (i.e. all
should be equal after some sequence of moves).
Input Specification:
The first line of the input contains one integer n (1 ≤ n ≤ 100) — the number of chips.
The second line of the input contains n integers
,
,…,
(1 ≤
≤ 109), where
is the coordinate of the i-th chip.
Output Specification:
Print one integer — the minimum total number of coins required to move all n chips to the same coordinate.
Sample Input1:
代码语言：javascript复制3
1 2 3Sample Output1:
代码语言：javascript复制1Sample Input2:
代码语言：javascript复制5
2 2 2 3 3Sample Output2:
代码语言：javascript复制2解题思路：
这道题目的大意说白了就是输出奇偶数中个数较少的那个。判断奇偶性的时候用n&1或者n%2都行，这俩个是等价的。
AC代码：
代码语言：javascript复制#include <bits/stdc .h>
using namespace std;
#define Up(i,a,b) for(int i = a; i <= b; i )
int main()
{
ios::sync_with_stdio(false);
int n, odd = 0, even = 0;
cin >> n;
Up(i,1,n)
{
int _;
cin >> _;
(_%2 ? odd : even ); //判断奇偶性，并记录奇偶数的个数
}
cout << min(odd,even) << endl;
return 0;
}
