Sonya and Robots（CodeForces 1004C）

Since Sonya is interested in robotics too, she decided to construct robots that will read and recognize numbers.

Sonya has drawn $\mathrm{nn numbers\; in\; a\; row, aiai is\; located\; in\; the ii-th\; position.\; She\; also\; has\; put\; a\; robot\; at\; each\; end\; of\; the\; row\; (to\; the\; left\; of\; the\; first\; number\; and\; to\; the\; right\; of\; the\; last\; number).\; Sonya\; will\; give\; a\; number\; to\; each\; robot\; (they\; can\; be\; either\; same\; or\; different)\; and\; run\; them.\; When\; a\; robot\; is\; running,\; it\; is\; moving\; toward\; to\; another\; robot,\; reading\; numbers\; in\; the\; row.\; When\; a\; robot\; is\; reading\; a\; number\; that\; is\; equal\; to\; the\; number\; that\; was\; given\; to\; that\; robot,\; it\; will\; turn\; off\; and\; stay\; in\; the\; same\; position.}$

Sonya does not want robots to break, so she will give such numbers that robots will stop before they meet. That is, the girl wants them to stop at different positions so that the first robot is to the left of the second one.

For example, if the numbers $[1,5,4,1,3][1,5,4,1,3] are\; written,\; and\; Sonya\; gives\; the\; number 11 to\; the\; first\; robot\; and\; the\; number 44 to\; the\; second\; one,\; the\; first\; robot\; will\; stop\; in\; the 11-st\; position\; while\; the\; second\; one\; in\; the 33-rd\; position.\; In\; that\; case,\; robots\; will\; not\; meet\; each\; other.\; As\; a\; result,\; robots\; will\; not\; be\; broken.\; But\; if\; Sonya\; gives\; the\; number 44 to\; the\; first\; robot\; and\; the\; number 55 to\; the\; second\; one,\; they\; will\; meet\; since\; the\; first\; robot\; will\; stop\; in\; the 33-rd\; position\; while\; the\; second\; one\; is\; in\; the 22-nd\; position.$

Sonya understands that it does not make sense to give a number that is not written in the row because a robot will not find this number and will meet the other robot.

Sonya is now interested in finding the number of different pairs that she can give to robots so that they will not meet. In other words, she wants to know the number of pairs ($\mathrm{pp, qq),\; where\; she\; will\; give pp to\; the\; first\; robot\; and qq to\; the\; second\; one.\; Pairs\; (pipi, qiqi)\; and\; (pjpj, qjqj)\; are\; different\; if pi\ne pjpi\ne pj or qi\ne qjqi\ne qj.}$

Unfortunately, Sonya is busy fixing robots that broke after a failed launch. That is why she is asking you to find the number of pairs that she can give to robots so that they will not meet.

Input

The first line contains a single integer $\mathrm{nn (1\le n\le 1051\le n\le 105) —\; the\; number\; of\; numbers\; in\; a\; row.}$

The second line contains $\mathrm{nn integers a1,a2,\dots ,ana1,a2,\dots ,an (1\le ai\le 1051\le ai\le 105) —\; the\; numbers\; in\; a\; row.}$

Output

Print one number — the number of possible pairs that Sonya can give to robots so that they will not meet.

Examples

input

Copy

5
1 5 4 1 3

output

Copy

9

input

Copy

7
1 2 1 1 1 3 2

output

Copy

7

Note

In the first example, Sonya can give pairs ($\mathrm{11, 11),\; (11, 33),\; (11, 44),\; (11, 55),\; (44, 11),\; (44, 33),\; (55, 11),\; (55, 33),\; and\; (55, 44).}$

In the second example, Sonya can give pairs ($\mathrm{11, 11),\; (11, 22),\; (11, 33),\; (22, 11),\; (22, 22),\; (22, 33),\; and\; (33, 22).}$

$\mathrm{题解：}$1-n:是每个机器人的排列顺序 ，然后组合对中(a, b)，b的位置不能在a的前面； 然后求这样的组合对一共有多少个； 逆向求解，dp[i]，表示i以后有多少个不同的数，注意int 会爆，用long long 解决。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cstring>
 5 #include<string>
 6 #include<cmath>
 7 #include<iomanip>
 8 #include<map>
 9 #include<stack>
10 #include<vector>
11 #include<queue>
12 #include<set>
13 #include<utility>
14 #include<list>
15 #include<algorithm>
16 #include <ctime>
17 #define max(a,b)   (a>b?a:b)
18 #define min(a,b)   (a<b?a:b)
19 #define swap(a,b)  (a=a+b,b=a-b,a=a-b)
20 #define memset(a,v)  memset(a,v,sizeof(a))
21 #define X (sqrt(5)+1)/2.0
22 #define maxn 320007
23 #define N 200005
24 #define INF 0x3f3f3f3f
25 #define PI acos(-1)
26 #define lowbit(x) (x&(-x))
27 #define read(x) scanf("%d",&x)
28 #define put(x) printf("%d\n",x)
29 #define memset(x,y) memset(x,y,sizeof(x))
30 #define Debug(x) cout<<x<<" "<<endl
31 #define lson i << 1,l,m
32 #define rson i << 1 | 1,m + 1,r
33 #define mod 1000000009
34 #define e  2.718281828459045
35 #define eps 1.0e-8
36 #define ll long long
37 using namespace std;
38 
39 
40 
41 ll dp[maxn];
42 int a[maxn], vis[maxn], vis2[maxn];
43 void init()
44 {
45     memset(dp, 0);
46     memset(a, 0);
47     memset(vis, 0);
48     memset(vis2, 0);
49 }
50 int main()
51 {
52     int n;
53     while (cin >> n)
54     {
55         init();
56         for (int i = 1; i <= n; i++)
57         {
58             cin >> a[i];
59         }
60         dp[n + 1] = 0;
61         for (int i = n; i >= 1; i--)//用数组下标处理，保证每个数只出现一次。
62         {
63             if (vis[a[i]] != 0)
64             {
65                 dp[i] = dp[i + 1];
66             }
67             else
68             {
69                 dp[i] = dp[i + 1] + 1;
70                 vis[a[i]] = 1;
71             }
72         }
73         ll ans = 0;
74         /*for(int i=1;i<=n;i++)
75             cout<<dp[a[i]]<<endl;*/
76         for (int i = 1; i <= n; i++)//将所有可能的条件找出
77         {
78             if (!vis2[a[i]])
79             {
80                 ans += dp[i + 1];
81                 vis2[a[i]] = 1;
82             }
83         }
84         cout << ans << endl;
85     }
86     return 0;
87 }

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