Codeforces round #261 (Div. 2) D tree Array application

[1, I] The number of medium to a [I] must be greater than [, JN] The number of medium to a [J] And I <J, number of such (I, j), I <J, but the valid number before I must be greater than the number after J, which looks like the number in reverse order, so it's easy to go To the tree array, but the processing is personal, as I encountered a lot of trouble in the game, although it took a lot of time, after jiege's education, it was actually possible to plug it into a tree array and look for it. It was very simple and clear.

int n;int aa[1000000 + 55];int bb[1000000 + 55];int c[1000000 + 55];map<int ,int >mp;ll lowbit(ll x) {return x&(-x);}void add(int i,int val) {while(i <= n) {c[i] += val;i += lowbit(i);}}ll get_sum(int i) {ll sum = 0;while(i) {sum += c[i];i -= lowbit(i);}return sum;}void init() {memset(c,0,sizeof(c));memset(aa,0,sizeof(aa));memset(bb,0,sizeof(bb));mp.clear();}int main() {while(scanf("%d",&n) == 1) {init();for(int i=1;i<=n;i++)scanf("%d",&aa[i]);for(int i=1;i<=n;i++) {mp[aa[i]]++;bb[i] = mp[aa[i]];add(bb[i],1);}mp.clear();ll ans = 0ll;for(int i=n;i>=1;i--) {add(bb[i],-1);mp[aa[i]]++;int tmp = mp[aa[i]];ans += i - get_sum(tmp) - 1;}cout<<ans<<endl;}return 0;}