Super OJ 序列计数

题意：

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。

分析：

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

代码：

#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;   
}

Super OJ 序列计数

原文：https://www.cnblogs.com/kma093/p/11791403.html