题目链接：Codeforces 459D Pashmak and Parmida‘s problem

题目大意：给定一个序列，定义f(l,r,x)为l≤k≤r并且ak=x的k的个数，求1≤i<j≤n的情况下，f(1,i,ai)<f(j,n,aj)的个数。

解题思路：预处理出f(1,i,ai)，f(j,n,aj)值，然后用树状数组维护个数。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
typedef long long ll;
const int maxn = 1e6+5;

struct state {
    int val, pos;
}sta[maxn];

bool cmp (const state& a, const state& b) {
    if (a.val != b.val)
        return a.val < b.val;
    return a.pos < b.pos;
}

int n, c[maxn], f[maxn];
ll v[maxn];

void add (int x, int val) {  
    while (x <= n) {  
        v[x] += val;  
        x += (x & (-x));  
    } 
} 

ll sum(int x) {  
    ll ans = 0;  
    while (x > 0) {  
        ans += v[x];  
        x -= (x &(-x));  
    } 
    return ans;  
}

ll solve () {
    ll ret = 0;
    memset(v, 0, sizeof(v));

    for (int i = n-1; i >= 0; i--) {
        ret += sum(c[i]-1);
        add(f[i], 1);
    }
    return ret;
}

int main () {
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        scanf("%d", &sta[i].val);
        sta[i].pos = i;
    }
    sort(sta, sta+n, cmp);

    int pre = -1, cnt = 0;
    for (int i = 0; i < n; i++) {
        if (pre != sta[i].val) {
            pre = sta[i].val;
            cnt = 1;
        } else
            cnt++;
        c[sta[i].pos] = cnt;
    }

    pre = -1, cnt = 0;
    for (int i = n; i >= 0; i--) {
        if (pre != sta[i].val) {
            pre = sta[i].val;
            cnt = 1;
        } else
            cnt++;
        f[sta[i].pos] = cnt;
    }

    printf("%lld\n", solve());
    return 0;
}