链接
https://www.luogu.com.cn/problem/P1637
题目
思路
事实上和求逆序对的题目有点像，但是求的是同序对（？。 先回顾下树状数组求逆序对的题目。 https://www.cnblogs.com/zzzsacmblog/p/18314521 这个总的思路其实就是前缀和，只不过拿树状数组优化了。先给每个节点对应的值对应树上的节点+1，然后求比它值小1的前缀和就是数量。代码
#include<iostream>
#include<vector>
#include<algorithm>
#include<math.h>
#include<sstream>
#include<string>
#include<string.h>
#include<iomanip>
#include<stdlib.h>
#include<map>
#include<queue>
#include<limits.h>
#include<climits>
#include<fstream>
#include<stack>
#define IOS ios::sync_with_stdio(false), cin.tie(0) ,cout.tie(0)
using namespace std;
typedef long long ll;
const int N = 3e4 + 10;
#define int long long
int a[N], b[N];
int tree[N << 2];
int lowbit(int x)
{
return x & (-x);
}
void add(int x)
{
int xnow = x;
while (xnow < N)
{
tree[xnow]++;
xnow += lowbit(xnow);
}
}
int sum(int x)
{
int ans = 0;
int xnow = x;
while (xnow > 0)
{
ans += tree[xnow];
xnow -= lowbit(xnow);
}
return ans;
}
stack<int>st; queue<int>q;
int suma[N];
signed main()
{
IOS;
int n; cin >> n;
for (int i = 1; i <= n; i++)
{
cin >> a[i];
b[i] = a[i];
}
sort(b + 1, b + 1 + n);
//stac
int cnt = 1;
map<int, int>small;
for (int i = 1; i <= n; i++)
{
if (!small[b[i]])
small[b[i]] = cnt++;
}
for (int i = 1; i <= n; i++)
a[i] = small[a[i]];
for (int i = n; i >= 1; i--)
{
add(a[i]);
st.push(n - i - sum(a[i]) + 1);
}
st.pop();
memset(tree, 0, sizeof(tree));
for (int i = 1; i <= n; i++)
{
add(a[i]);
q.push(sum(a[i]-1));
}
q.pop();
ll ans = 0;
for (int i = 2; i < n; i++)
{
ans += q.front() * st.top();
q.pop(), st.pop();
}
cout << ans;
return 0;
}