Skip to main content

Cleaning Tables- CLETAB

Problem Link

  1. #include <iostream>
  2. #include <cstring>
  3. #include <algorithm>
  4.  
  5. using namespace std;
  6.  
  7. int main()
  8. {
  9. int t, n, m, res, victim;
  10. int c[401];
  11. int lastUse[401];
  12. int seenBefore[401];
  13. bool present[401];
  14.  
  15. cin >> t;
  16. while(t--)
  17. {
  18. cin >> n >> m;
  19. memset(present, false, sizeof present);
  20. memset(seenBefore, 0, sizeof seenBefore);
  21.  
  22. for(int i = 0; i < m; i++)
  23. {
  24. cin >> c[i];
  25. lastUse[c[i]] = i;
  26. }
  27.  
  28. int order = 0;
  29. res = 0;
  30. while((res < n || present[c[order]]) && order < m)
  31. {
  32. if(!present[c[order]])
  33. {
  34. res++;
  35. present[c[order]] = true;
  36. }
  37.  
  38. order++;
  39. }
  40.  
  41. while(order < m)
  42. {
  43. if(!present[c[order]])
  44. {
  45. victim = -1;
  46.  
  47. for(int i = order - 1; i >= 0; i--)
  48. {
  49. if(present[c[i]] && lastUse[c[i]] <= order)
  50. {
  51. victim = c[i];
  52. break;
  53. }
  54. }
  55.  
  56. if(victim == -1)
  57. {
  58. for(int i = order + 1; i < m; i++)
  59. {
  60. if(present[c[i]] && seenBefore[c[i]] != order)
  61. {
  62. victim = c[i];
  63. seenBefore[c[i]] = order;
  64. }
  65. }
  66. }
  67.  
  68. present[victim] = false;
  69. present[c[order]] = true;
  70.  
  71. res++;
  72. }
  73.  
  74. order++;
  75. }
  76.  
  77. cout << res << endl;
  78.  
  79. }
  80. return 0;
  81. }

Comments

Popular posts from this blog

GCJ101BB - Picking Up Chicks

Problem Link /* explanation     lets solve the problem only for 2 chicken.     s[i] = speed of chicken i     pos[i] = position of chicken i     if s[i] > s[i - 1] then no problem, just check whether both can reach b within time or not.     if s[i] < s[i - 1] then there is a chance that i can slow down i - 1.     lets say s[i] = 1 m/sec and s[i - 1] = 2 m/sec and time limit is T and point to reach is B.     for s[i] pos[i] can be at max B - T. if pos is greater than B-T it can not reach within Tsec.     and for s[i - 1] pos[i - 1] can be at max (B-T)*2. if pos[i - 1] > (B-T)*2 it can not reach within Tsec.     at T sec i will be at B and i - 1 will also be at B. at T - 1 i will be at B-T-1 and i-1 will be at B-T-2 and so on. as we can see i -1 will always be behind i. so there will not be any collision.     if i is pos[i] < B-T then i can reach B before T sec ...

KOPC12A

KOPC12A - K12 - Building Construction #include <iostream> #include <cmath> #define REP(i, n) for(int i = 0; i < n; i++) using namespace std; const int N = 10005; int height[N]; int cost[N]; int n; //finds the total cost for height h long long int findCost(int h) {     long long c = 0; //cost     REP(i, n)     {         c += abs(h - height[i]) * cost[i];     }     return c; } int ternary_search(int l, int h) {     while(l <= h)     {         if(l == h)             break;         int mid1 = l + (h - l) / 3;         int mid2 = h - (h - l) / 3;         if(findCost(mid1) > findCost(mid2))             l = mid1 + 1;         else ...

Yet Another Cute Girl - PRETNUM

Problem Link #include <iostream> #include <cstring> #include <cmath> #include <vector> using namespace std; #define LL long long const int N = 1000010; bool prime[N + 1]; vector<int> v; void sieve() { for(int i = 2; i <= N; i++) { prime[i] = true; } for(int i = 2; i * i <= N; i++) { if(prime[i]) { for(int j = i * i; j <= N; j += i) { prime[j] = false; } } } v.push_back(2); for(int i = 3; i <= N; i += 2) { if(prime[i]) { v.push_back(i); } } } LL getNumDiv(LL num) {     /* if prime factor of n is p1^k1 p2 ^ k2 then prime factor of n^2 would be         (p1^k1 p2 ^ k2)^2 = p1^(2k1) p2^(2k2)         so number of divisors of n^2 would be (2 * k1 + 1) * (2 * k2 + 1) ...         if n is a prime number number then number of divisors of n^2 would be 3.     */ LL res = 1; for(int i = 0;...