题目链接：逆序数

解题思路：逆序数模板题，将数值离散化后把求值的逆序数转化成求其rank的逆序数。利用树状数组提升标记数组的效率。注意i - get_sum(rnk[i])表示在长度为i，或者说第i个数，之前排名比他大数量。

AC代码：

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <string>
#include <deque>
#include <queue>
#include <set>
#include <map>
#include <stack>
#include <vector>
#include <utility>

using namespace std;
const int maxn = (int)1e5+5;
typedef long long ll;

ll sum;
int c[maxn],val[maxn],tmp[maxn],rnk[maxn],n;

inline int lowbit (int a) {
	return a & (-a);
}

inline void add (int idx, int num) {
	while (idx <= n) {
		c[idx] += num;
		idx += lowbit(idx);
	}
}

inline ll get_sum (int idx) {
	ll ans = 0;
	while (idx > 0) {
		ans += c[idx];
		idx -= lowbit(idx);
	}
	return ans;
}

void solve() {
	//离散化
	for (int i = 1; i <= n; i++) {
		cin >> val[i];
		tmp[i] = val[i];
	}
	sort(tmp + 1, tmp + 1 + n);
	int len = unique(tmp + 1, tmp + 1 + n) - tmp - 1;
	for (int i = 1; i <= n; i++) {
		rnk[i] = lower_bound(tmp + 1, tmp + 1 + len, val[i]) - tmp; //rnk [1,...]
	}
	
	//从比较val的值转化成维护其rnk 找rnk逆序数
	memset(c, 0, sizeof(c));
	sum = 0;
	for (int i = 1; i <= n; i++) {
		add(rnk[i], 1);
		sum += i - get_sum(rnk[i]);
	}
	
	cout << sum << '\n';
}

int main() {
	ios::sync_with_stdio(false);
	cin >> n;
	solve();
	return 0;
}