CF - 主席树 E

Posted 2020-10-31

One day Polycarp decided to rewatch his absolute favourite episode of well-known TV series "Tufurama". He was pretty surprised when he got results only for season 7 episode 3 with his search query of "Watch Tufurama season 3 episode 7 online full hd free". This got Polycarp confused — what if he decides to rewatch the entire series someday and won‘t be able to find the right episodes to watch? Polycarp now wants to count the number of times he will be forced to search for an episode using some different method.

TV series have n seasons (numbered 1 through n), the i-th season has a_i episodes (numbered 1 through a_i). Polycarp thinks that if for some pair of integers x and y (x?<?y) exist both season x episode y and season y episode x then one of these search queries will include the wrong results. Help Polycarp to calculate the number of such pairs!

Input

The first line contains one integer n (1??≤?n??≤??2·10⁵) — the number of seasons.

The second line contains n integers separated by space a₁,?a₂,?...,?a_n (1?≤?a_i?≤?10⁹) — number of episodes in each season.

Output

Print one integer — the number of pairs x and y (x?<?y) such that there exist both season x episode y and season y episode x.

Examples

Input

Copy

5
1 2 3 4 5

Output

Copy

0

Input

Copy

3
8 12 7

Output

Copy

3

Input

Copy

3
3 2 1

Output

Copy

2

Note

Possible pairs in the second example:

1. x?=?1, y?=?2 (season 1 episode 2 season 2 episode 1);
2. x?=?2, y?=?3 (season 2 episode 3 season 3 episode 2);
3. x?=?1, y?=?3 (season 1 episode 3 season 3 episode 1).

In the third example:

1. x?=?1, y?=?2 (season 1 episode 2 season 2 episode 1);
2. x?=?1, y?=?3 (season 1 episode 3 season 3 episode 1).

 题意 ： 询问存在多少个点对，使得 a[j] >= i && a[i] >= j 

两种方法 ：

一 ： 主席树，这个比较好想，首先我们比较容易保证的条件是 a[j] >= i , 然后对于这个区间我们去寻找有多少个点是 a[i] >= j, 那么问题不就转变成求一个指定区间比一个指定的数大的个数。

代码示例 ：

#define ll long long
const ll maxn = 2e5+5;
const ll mod = 1e9+7;
const double eps = 1e-9;
const double pi = acos(-1.0);
const ll inf = 0x3f3f3f3f;

ll n;
ll a[maxn];
ll root[maxn];
ll cnt = 1;
struct node
{
    ll l, r;
    ll sum;
}t[maxn*20];
void init(){
    root[0] = 0;
    t[0].l = t[0].r = t[0].sum = 0;
}

void update(ll num, ll &rt, ll l, ll r){
    t[cnt++] = t[rt];
    rt = cnt-1;
    t[rt].sum++;
    if (l == r) return;
    ll m = (l+r)>>1;
    if (num <= m) update(num, t[rt].l, l, m);
    else update(num, t[rt].r, m+1, r);
}

ll sum = 0;
void query(ll i, ll j, ll k, ll l, ll r){ 
    if (l >= k) {
        sum += t[j].sum-t[i].sum;
        return;
    } 
    ll m = (l+r)>>1;
    if (k <= m) query(t[i].l, t[j].l, k, l, m);
    query(t[i].r, t[j].r, k, m+1, r);
    
}

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    cin >> n;
    for(ll i = 1; i <= n; i++){
        scanf("%lld", &a[i]);
        a[i] = min(a[i], n);
    }
    for(ll i = 1; i <= n; i++){
        root[i] = root[i-1];
        update(a[i], root[i], 1, n);
    }
    
    for(ll i = 1; i <= n; i++){
        ll l = i+1, r = a[i];
        if (l > r) continue;
        query(root[l-1], root[r], i, 1, n);
    }
    printf("%lld\n", sum);
    return 0;
}

 方法二 ： 树状数组