给出序列 a1，a2，……an（0≤ai≤109），求三元组（ai，aj，ak）（1≤i＜j＜k≤n）满足 ai＜aj＞ak 的数量。
开两个\(BIT\)，分别维护前面比它小的和后面比它大的，然后组合计数一下即可
#include<bits/stdc++.h> #define lowbit(x) (x & (-x)) #define ll long long #define N (100000 + 5) using namespace std; inline int read() { int cnt = 0, f = 1; char c = getchar(); while (!isdigit(c)) {if (c == '-') f = -f; c = getchar();} while (isdigit(c)) {cnt = (cnt << 3) + (cnt << 1) + c - '0'; c = getchar();} return cnt * f; } int n, q, a[N], b[N << 1]; ll ans; void pre() { sort(b + 1, b + n + 1); q = unique(b + 1, b + n + 1) - b - 1; for (register int i = 1; i <= n; ++i) a[i] = lower_bound(b + 1, b + q + 1, a[i]) - b; } struct node { int BIT[N]; void insert (int x) {for (; x <= n; x += lowbit(x)) ++BIT[x];} void Delete (int x) {for (; x <= n; x += lowbit(x)) --BIT[x];} ll query(int x) {ll ans = 0; for (; x; x -= lowbit(x)) ans += BIT[x]; return ans;} }BIT1, BIT2; int main() { n = read(); for (register int i = 1; i <= n; ++i) a[i] = b[i] = read(); pre(); for (register int i = 1; i <= n; ++i) BIT2.insert(a[i]); BIT1.insert(a[1]); for (register int i = 2; i <= n; ++i) { BIT1.insert(a[i]); BIT2.Delete(a[i - 1]); ans += (BIT1.query(a[i] - 1) * (BIT2.query(a[i] - 1))); // cout<<BIT1.query(a[i] - 1)<< " " << BIT2.query(a[i])<<"\n"; } printf("%lld", ans); return 0; } 来源：博客园
作者：kma_093
链接：https://www.cnblogs.com/kma093/p/11791403.html