codeforces ~ 1004 C Sonya and Robots (dp)

C. Sonya and Robots

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 ($\mathrm{11, 11),\; (11, 22),\; (11, 33),\; (22, 11),\; (22, 22),\; (22, 33),\; and\; (33, 22).}$

题意：

给你n个数字，有两个不同的机器人，一个在数列的最左端，另一个在数列的最右端，

如果机器人1标号为x，那么他从左往右会在第一个x这里停下来

第二个机器人标号为y，他会从右往左在第一个y这里停下来

求（x，y）这样的一共有多少对，可以使得机器人不相撞

题解：

左边的机器人遇见重复的就不能往后走了

右边的机器人遇见重复的就不能往前走了

那么我们先将前n个数中不重复的表给打出来

那么从右往左走的时候，不同的个数就是我们要求的对数

注意开longlong

代码如下

#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5+5;
int a[maxn];
int vis[maxn];
int dp[maxn];
int main(){
  int n;
  while(cin>>n){
    memset(vis,0,sizeof(vis));
    memset(dp,0,sizeof(dp));
    dp[0]=0;
    for(int i=1;i<=n;i++){
      cin>>a[i];
      if(!vis[a[i]]){
        vis[a[i]]=1;
        dp[i]=dp[i-1]+1;
      }else{
        dp[i]=dp[i-1];
      }
    }
    long long ans=0;
    memset(vis,0,sizeof(vis));
    for(int i=n;i>=1;i--){
      if(!vis[a[i]]){
        vis[a[i]]=1;
        ans+=dp[i-1];
      }
    }
    printf("%I64d\n",ans);

  }
}

View Code