Educational Codeforces Round 41 (Rated for Div. 2) E. Tufurama

Posted 2020-11-11 hit_yjl

E. Tufurama

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 ai episodes (numbered 1 through ai). 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,?...,?an (1?≤?ai?≤?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).

思路一：对于每一个a[i], 我们要统计的是[i + 1, a[i] ]区间内有多少个a[j] >= i，于是就是一个很显然的线段树套pbds，时间复杂度nlog2n, 空间复杂度nlogn。

 1 #include <iostream>
 2 #include <fstream>
 3 #include <sstream>
 4 #include <cstdlib>
 5 #include <cstdio>
 6 #include <cmath>
 7 #include <string>
 8 #include <cstring>
 9 #include <algorithm>
10 #include <queue>
11 #include <stack>
12 #include <vector>
13 #include <set>
14 #include <map>
15 #include <list>
16 #include <iomanip>
17 #include <cctype>
18 #include <cassert>
19 #include <bitset>
20 #include <ctime>
21 
22 using namespace std;
23 
24 #define ll long long
25 #define pii pair<int, int>
26 #define pb push_back
27 #define mp make_pair
28 #define clr(a, x) memset(a, x, sizeof(a))
29 
30 const double pi = acos(-1.0);
31 const int INF = 0x3f3f3f3f;
32 const int MOD = 1e9 + 7;
33 const double EPS = 1e-9;
34 
35 
36 #include <ext/pb_ds/assoc_container.hpp>
37 #include <ext/pb_ds/tree_policy.hpp>
38 
39 using namespace __gnu_pbds;
40 
41 tree<pii, null_type, greater<pii>, rb_tree_tag, tree_order_statistics_node_update> T[800015];
42 int n, a[200015];
43 
44 void build(int i, int l, int r) {
45     for (int j = l; j <= r; ++j) {
46         T[i].insert(pii(a[j], j));
47     }
48     if (l == r) {
49         return;
50     }
51     int mi = l + r >> 1;
52     build(i << 1, l, mi);
53     build(i << 1 | 1, mi + 1, r);
54 }
55 int query(int i, int l, int r, int x, int y, int z) {
56     if (x <= l && r <= y) {
57         return T[i].order_of_key(pii(z, 0));
58     }
59     int mi = l + r >> 1;
60     int res = 0;
61     if (x <= mi) res += query(i << 1, l, mi, x, y, z);
62     if (mi < y) res += query(i << 1 | 1, mi + 1, r, x, y, z);
63     return res;
64 }
65 int main() {
66     scanf("%d", &n);
67     for (int i = 1; i <= n; ++i) {
68         scanf("%d", &a[i]);
69     }
70     build(1, 1, n);
71     ll res = 0;
72     for (int i = 1; i <= n; ++i) {
73         if (i < a[i]) {
74             res += query(1, 1, n, i + 1, min(a[i], n), i);
75         }
76     }
77     printf("%lld\n", res);
78     return 0;
79 }

View Code

思路二：我们将每个值向后的影响称为种韭菜，将统计每个值之前有影响的点对称为割韭菜。我们不按照下标的顺序割韭菜，而是处理一下，对于每个点，若a[i] >= i, 则之前种的韭菜都能收割到，割韭菜的时机还是i；否则割韭菜的时机就要改为a[i]。所以最终就树状数组正常更新就行了，只是更改一下查询的时间点。代码比思路一快了10倍。

 1 #include <iostream>
 2 #include <fstream>
 3 #include <sstream>
 4 #include <cstdlib>
 5 #include <cstdio>
 6 #include <cmath>
 7 #include <string>
 8 #include <cstring>
 9 #include <algorithm>
10 #include <queue>
11 #include <stack>
12 #include <vector>
13 #include <set>
14 #include <map>
15 #include <list>
16 #include <iomanip>
17 #include <cctype>
18 #include <cassert>
19 #include <bitset>
20 #include <ctime>
21 
22 using namespace std;
23 
24 #define pau system("pause")
25 #define ll long long
26 #define pii pair<int, int>
27 #define pb push_back
28 #define mp make_pair
29 #define clr(a, x) memset(a, x, sizeof(a))
30 
31 const double pi = acos(-1.0);
32 const int INF = 0x3f3f3f3f;
33 const int MOD = 1e9 + 7;
34 const double EPS = 1e-9;
35 
36 /*
37 #include <ext/pb_ds/assoc_container.hpp>
38 #include <ext/pb_ds/tree_policy.hpp>
39 
40 using namespace __gnu_pbds;
41 tree<pli, null_type, greater<pli>, rb_tree_tag, tree_order_statistics_node_update> T;
42 */
43 
44 int n, a[200015], c[200015];
45 vector<int> q[200015];
46 void add(int p, int x) {for (; p <= n; p += p & -p) c[p] += x;}
47 int query(int p) {
48     int res = 0;
49     for (; p; p -= p & -p) res += c[p];
50     return res;
51 }
52 int main() {
53     scanf("%d", &n);
54     for (int i = 1; i <= n; ++i) {
55         scanf("%d", &a[i]);
56         if (a[i] > i) q[i].pb(i);
57         else q[a[i]].pb(i);
58     }
59     ll ans = 0;
60     for (int i = 1; i <= n; ++i) {
61         if (i < a[i]) {
62             add(i + 1, 1);
63             add(a[i] + 1, -1);
64         }
65         for (int j = 0; j < q[i].size(); ++j) {
66             ans += query(q[i][j]);
67         }
68     }
69     printf("%lld\n", ans);
70     return 0;
71 }

View Code