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/*
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树状数组解法 https://blog.csdn.net/m0_50299150/article/details/122729602
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权值线段树解法
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*/
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#include <bits/stdc++.h>
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using namespace std;
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typedef long long LL;
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const int N = 5e5 + 10;
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int a[N], b[N];
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int n;
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// 344 ms
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struct Node {
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int l, r; //描述的范围[l,r]
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int sum; // l,r之间数字的个数
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} tr[N << 2]; // 线段树需要开4倍的空间
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void build(int u, int l, int r) {
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tr[u] = {l, r, 0};
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if (l == r) return;
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int mid = (l + r) >> 1;
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build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
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}
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void pushup(int u) {
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tr[u].sum = tr[u << 1].sum + tr[u << 1 | 1].sum;
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}
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void modify(int u, int p, int v) {
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if (tr[u].l == p && tr[u].r == p) {
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tr[u].sum += v;
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return;
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}
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int mid = tr[u].l + tr[u].r >> 1;
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if (p <= mid)
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modify(u << 1, p, v);
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else
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modify(u << 1 | 1, p, v);
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//更新父节点信息
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pushup(u);
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}
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int query(int u, int l, int r) {
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if (tr[u].l == l && tr[u].r == r) return tr[u].sum;
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int mid = tr[u].l + tr[u].r >> 1;
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if (mid < l)
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return query(u << 1 | 1, l, r);
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else {
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if (r <= mid)
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return query(u << 1, l, r);
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else
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return query(u << 1, l, mid) + query(u << 1 | 1, mid + 1, r);
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}
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}
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int main() {
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//本题，如果使用vector+离散化，在AcWing可以正常AC，但在洛谷会有很多TLE，目前想来原因可能是因为vector的性能导致
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//如果采用静态的数组进行离散化处理，就不会有这个问题
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cin >> n;
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//一般我们使用线段树，都喜欢从下标1开始，所以我们在处理数据时尽量让下标从1开始
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for (int i = 1; i <= n; i++) cin >> a[i], b[i] = a[i];
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sort(a + 1, a + n + 1); //对原始数组进行排序
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//这里没有进行离散化，只是对数据重新规范到范围[1,n]之间
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for (int i = 1; i <= n; i++) b[i] = lower_bound(a + 1, a + n + 1, b[i]) - a;
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LL ans = 0;
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//创建线段树
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build(1, 1, n + 1);
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//边查边添加,查找它后面比它大的数字个数
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for (int i = 1; i <= n; i++) {
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ans += query(1, b[i] + 1, n + 1);
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modify(1, b[i], 1);
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}
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//输出答案
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cout << ans << endl;
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return 0;
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}
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