Codeforces Round #495 (Div. 2)(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.

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

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

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

$\mathrm{题目大意:有n个数,每个数只能与自己后面的数配对,相同的配对只算一个,求配对的数量.}$

$\mathrm{分析：我们可以倒着往前压，把数放进map(去重操作）中，从后往前放入容器中，ans[i]就等于容器中元素的个数，并且我们用mp[str[i]]=i,记录该元素第一次出现的位置以方便后面的更新，若该元素第二次出现，上一次出现的位置ans[pos]=0,因为位置越靠前，配对的个数越多}$

 1 #include <iostream>
 2 #include<cmath>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<stack>
 6 #include<queue>
 7 #include<deque>
 8 #include<map>
 9 #include<algorithm>
10 #define PI acos(-1.0)
11 using namespace std;
12 typedef long long ll;
13 map<int,int>::iterator it;
14 int str[111234];
15 int ans[123114];
16 int m,n;
17 int main()
18 {
19     scanf("%d",&m);
20     map<int,int>mp;
21     mp.clear();
22     for(int i=1;i<=m;i++)
23     {
24         scanf("%d",&str[i]);
25     }
26     for(int i=m;i>=1;i--)
27     {
28         ans[i]=mp.size();
29         if(mp.count(str[i]))//更新操作，若该元素第二次出现，上一次的ans赋值为0
30         {
31             ans[mp[str[i]]]=0;
32         }
33         mp[str[i]]=i;
34     }
35     ans[m+1]=‘\0‘;
36     ll sum=0;
37      for(int i=1;i<=m;i++)
38         sum+=ans[i];
39         printf("%lld\n",sum);
40      return 0;
41 
42 }