Created
July 13, 2017 01:20
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Solução OBI 2017
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| // Ivan Carvalho | |
| // Cortando o Papel - Fase 2 Programação Nível 2 - OBI 2017 | |
| // O(n*log(n)) | |
| #include <bits/stdc++.h> | |
| using namespace std; | |
| typedef pair<int,int> ii; | |
| vector<ii> papel; | |
| int estado[100010],n,tiras,resp; | |
| int main(){ | |
| scanf("%d",&n); | |
| tiras = 1; | |
| resp = 1; | |
| for(int i=1;i<=n;i++){ | |
| estado[i] = 1; | |
| int hi; | |
| scanf("%d",&hi); | |
| papel.push_back(ii(hi,i)); | |
| } | |
| sort(papel.begin(),papel.end()); | |
| for(int i=0;i<n;i++){ | |
| int pos = papel[i].second; | |
| estado[pos] = 0; | |
| if(estado[pos-1] == 1 && estado[pos+1] == 1) tiras++; | |
| if(estado[pos-1] == 0 && estado[pos+1] == 0) tiras--; | |
| if(i == n-1 || papel[i].first != papel[i+1].first){ | |
| resp = max(resp,tiras); | |
| //printf("C %d\n",tiras); | |
| } | |
| } | |
| printf("%d\n",resp+1); | |
| return 0; | |
| } |
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