Codeforces 459D Pashmak and Parmida's problem(樹狀數組)，codeforces459d

Codeforces 459D Pashmak and Parmida's problem(樹狀數組)，codeforces459d

題目連結：Codeforces 459D Pashmak and Parmida's problem

題目大意：給定一個序列，定義f(l,r,x)為l≤k≤r並且ak=x的k的個數，求1≤i<j≤n的情況下，f(1,i,ai)<f(j,n,aj)的個數。

解題思路：預先處理出f(1,i,ai)，f(j,n,aj)值，然後用樹狀數組維護個數。

#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;}