Submission #313342

#TimeUsernameProblemLanguageResultExecution timeMemory
313342mohamedsobhi777Sure Bet (CEOI17_sure)C++14
60 / 100
2080 ms3736 KiB
#include<bits/stdc++.h> #pragma GCC optimize("-Ofast") //#pragma GCC optimize("trapv") #pragma GCC target("sse,sse2,sse3,ssse3,sse4,sse4.2,popcnt,abm,mmx,avx2,tune=native") #pragma GCC optimize("-ffast-math") #pragma GCC optimize("-funroll-loops") #define I inline void #define S struct #define vi vector<int> #define vii vector<pair<int,int>> #define pii pair<int,int> #define pll pair<ll,ll> using namespace std ; using ll = long long ; using ld = long double ; const int N = 2e5 + 7 , mod = 1e9 + 7 ; const ll inf = 2e18 ; // How interesting! int n ; vector<double> d[2], pd[2]; double ans ; void solve2(){ double sum1 = 0.0 ; for(int i = 0 ;i < n; ++ i){ sum1 += d[0][i] ; for(int j = 0 ;j < n ; ++ j){ int k = lower_bound(pd[1].begin() , pd[1].end() , sum1 + i + 1 ) - pd[1].begin() ; if(k != pd[1].size()){ ans = max(ans , sum1 - k - 1) ; } } } } int main(){ ios_base::sync_with_stdio(0) ; cin.tie(0) ; //freopen("in.in" , "r" , stdin) ; cin >> n; for(int i = 0 ;i < n; ++ i){ double a , b; cin >> a >> b ; a-=1.0 ; b-=1.0 ; d[0].push_back(a) ; d[1].push_back(b) ; } for(int i = 0 ;i < 2 ; ++ i){ sort(d[i].begin() , d[i].end(), greater<double>()) ; pd[i].resize(n) ; } pd[0][0] = d[0][0] +1.0; pd[1][0] = d[1][0] +1.0; for(int i = 0 ; i < 2 ; ++ i){ for(int j = 1 ;j < n ; ++ j) pd[i][j] = pd[i][j-1] + d[i][j] +1.0; } solve2() ; swap(d[0] , d[1]) ; swap(pd[0] , pd[1]) ; solve2() ; cout<< fixed <<setprecision(4) << ans ; return 0 ; } /* - bounds sir (segtree = 4N, eulerTour = 2N, ...) - a variable defined twice? - will overflow? - is it a good complexity? - don't mess up indices (0-indexed vs 1-indexed) - reset everything between testcases. */

Compilation message (stderr)

sure.cpp: In function 'void solve2()':
sure.cpp:36:30: warning: comparison of integer expressions of different signedness: 'int' and 'std::vector<double>::size_type' {aka 'long unsigned int'} [-Wsign-compare]
   36 |                         if(k != pd[1].size()){
      |                            ~~^~~~~~~~~~~~~~~
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