IQOverload

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Aug 31st, 2014
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  1. #include <algorithm>
  2. #include <bitset>
  3. #include <deque>
  4. #include <cmath>
  5. #include <cstdio>
  6. #include <cstdlib>
  7. #include <cstring>
  8. #include <iostream>
  9. #include <list>
  10. #include <map>
  11. #include <queue>
  12. #include <set>
  13. #include <sstream>
  14. #include <stack>
  15. #include <string>
  16. #include <utility>
  17. #include <vector>
  18.  
  19. #define fst first
  20. #define snd second
  21. #define all(x) (x).begin(), (x).end()
  22. #define clr( a , v ) memset( a , v , sizeof(a) )
  23. #define pb push_back
  24. #define mp make_pair
  25. #define sz size()
  26. #define FORN( i , s , n ) for( int64 i = s ; i < (int64)(n) ; i++ )
  27. #define FOR( i , n ) FORN( i , 0 , n )
  28. #define FORIT(i,x) for( typeof x.begin() i = x.begin() ; i != x.end() ; i++ )
  29. #define trace(x)    cout << #x << ": " << x << endl;
  30. #define trace2(x, y) cout << #x << ": " << x << " | " << #y << ": " << y << endl;
  31. #define read ios::sync_with_stdio(false)
  32.  
  33. using namespace std;
  34.  
  35. typedef long long int64;
  36.  
  37. const int64 N = 1000001;
  38. const int64 LN = 30; //log2(N) + 1
  39.  
  40. int64 a[N], Log2[N], p[LN][N];
  41.  
  42. void pre(){
  43.     int64 k = 0;
  44.     FORN( i , 1 , N+1 ){
  45.         Log2[i] = k;
  46.         if ( 1<<(k+1) == i ) k++;
  47.     }
  48. }
  49.  
  50. void init( int64 n ) {
  51.     int64 ln = Log2[n], i1, i2;
  52.     FOR( i , n ) p[0][i] = i;
  53.     FORN( i , 1 , ln+1 ){
  54.         FORN( j , 0 , n - (1<<i) + 1 ){
  55.             i1 = p[i-1][ j ];
  56.             i2 = p[i-1][ j + (1 << i-1) ];
  57.             p[i][j] = a[i1] <= a[i2] ? i1 : i2;
  58.         }
  59.     }
  60. }
  61.  
  62. int64 query(int64 i, int64 j) {
  63.         if( i > j ) return 0;
  64.     int64 ln = Log2[ j - i + 1 ];
  65.     int64 i1 = p[ ln ][ i ];
  66.     int64 i2 = p[ ln ][ j - (1 << ln) + 1 ];
  67.     return a[i1] <= a[i2] ? i1 : i2;
  68. }
  69.  
  70. int64 le[N], ri[N], acum[N];
  71.  
  72. int main(){
  73.     int64 n; cin >> n;
  74.     FOR( i , n ){
  75.         cin >> a[i];
  76.         if( i == 0 ) acum[i] = a[0];
  77.         else acum[i] = acum[i-1] + a[i];
  78.     }
  79.  
  80.     pre(); init( n+5 );
  81.  
  82.     FOR( i , n ){
  83.  
  84.         int64 hi = n-1, lo = i, mid, id;
  85.         while( hi - lo > 1 ){
  86.             mid = ( hi + lo ) / 2;
  87.             id = query( i , mid );
  88.             if( a[id] >= a[i] ) lo = mid;
  89.             else hi = mid;
  90.         }
  91.         id = query( i , hi );
  92.         if( a[id] >= a[i] ) ri[i] = hi;
  93.         else ri[i] = lo;
  94.        
  95.         hi = i, lo = 0;
  96.         while( hi - lo > 1 ){
  97.             mid = ( hi + lo ) / 2;
  98.             id = query( mid , i );
  99.             if( a[id] >= a[i] ) hi = mid;
  100.             else lo = mid;
  101.         }
  102.         id = query( lo , i );
  103.         if( a[id] >= a[i] ) le[i] = lo;
  104.         else le[i] = hi;   
  105.  
  106.     }
  107.  
  108.     int64 ans = 0;
  109.     int64 l, r, ansl, ansr, cur;
  110.     FOR( i , n ){
  111.         l = le[i], r = ri[i];
  112.         if( l == 0 ) cur = a[i] * acum[r];
  113.         else cur = a[i] * ( acum[r] - acum[l-1] );
  114.         if( ans <= cur ){
  115.             ans = cur;
  116.             ansl = l, ansr = r;
  117.         }
  118.     }
  119.     cout << ans << endl;
  120.     cout << ansl+1 << " " << ansr+1 << endl;
  121.     return 0;
  122. }
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