逆序对

求逆序对

1.

void merge_sort(int l, int r) {
	if(l == r) return;
	int mid = (l + r) / 2;
	merge_sort(l, mid);
	merge_sort(mid + 1, r);
	
	int i = l, j = mid + 1;
	int tem[MAXN], len = l;
	while(i <= mid && j <= r) {
		if(a[i] <= a[j]) {
			tem[len++] = a[i++];
		}
		else {
			tem[len++] = a[j++];
			ans += mid - i + 1;
		}
	}
	while(i <= mid) tem[len++] = a[i++];
	while(j <= r) tem[len++] = a[j++];
	for(int i = l; i <= r; i++) a[i] = tem[i];
	return;
}

2.

#include <iostream>
#include <algorithm>
using namespace std;

const int MAXN = 1005;
int n;
int rev[MAXN], a[MAXN], BIT[MAXN];

int lowbit(int x) {return x & -x;}
void _Update(int, int);
int _Sum(int);

int main() {
	cin >> n;
	for(int i = 1; i <= n; i++) cin >> a[i], rev[i] = a[i];
	sort(rev + 1, rev + 1 + n);
	int len = unique(rev + 1, rev + 1 + n) - rev - 1;
	for(int i = 1; i <= n; i++) {
		a[i] = lower_bound(rev + 1, rev + 1 + len, a[i]) - rev;
	}
	//离散化 
	
	int ans = 0;
	for(int i = 1; i <= n; i++) {
		_Update(a[i], 1);
		ans += i - _Sum(a[i]);
	}
	cout << ans;
	return 0;
}

void _Update(int index, int x) {
	for(int i = index; i <= n; i += lowbit(i))
		BIT[i] += x;
	return;
}

int _Sum(int index) {
	int sum = 0;
	for(int i = index; i > 0; i -= lowbit(i))
		sum += BIT[i];
	return sum;
}