Submission #4262331

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Source Code Expand

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <bitset>
#include <cmath>
#include <queue>
#include <map>
#include <set>
using namespace std;
#define rg register
#define LL long long
#define __space putchar(' ')
#define __endl putchar('\n')
template <typename qwq> inline void read(qwq & x)
{
    x = 0;
    rg int f = 1;
    rg char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-') f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
    {
        x = (x << 1) + (x << 3) + (c ^ 48);
        c = getchar();
    }
    x *= f;
}
template <typename qaq> inline void print(qaq x)
{
    if (x < 0)
    {
        putchar('-');
        x = -x;
    }
    if (x > 9) print(x / 10);
    putchar(x % 10 + '0');
}
struct node
{
	int val,data;
}a[233333];
int n;
LL ans;
bitset<1000000005> s;
inline bool cmp(node x,node y)//预处理为有序 
{
	return x.val < y.val;
}
int main()
{
	read(n);
	ans = n;
	for (rg int i = 1;i <= n;++i)
	{
		read(a[i].val);
		a[i].data = i;
	}
	stable_sort(a + 1,a + 1 + n,cmp);
	for (rg int i = 1;i <= n;++i)
	{
		if (s[a[i].val]) continue;
		rg bool odd_i = i & 1,odd_data = a[i].data & 1;
		if (odd_i == odd_data) s[a[i].val] = true,--ans;//若位置变化后与变化前的奇偶性相同
		//说明他们必须用操作1才可达到,其余的均可用操作2达到 
	}
	print(ans >> 1);//一个奇数位置需要操作1,定有一个对应的偶数位置需要操作1,故/ 2 
}

Submission Info

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