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Aug 25th, 2019
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  1. #include <bits/stdc++.h>
  2.  
  3. using namespace std;
  4.  
  5. vector<string> split_string(string);
  6.  
  7. // Complete the luckBalance function below.
  8. int luckBalance(int k, vector<vector<int>> contests) {
  9.  
  10. int out;
  11. cout << contests[0][0] << contests [0][1] << endl;
  12.  
  13. int imp = 0;
  14. //cout << contests.size() << endl; 6
  15. int sum = 0;
  16. vector<int> vect(contests.size(), 0);
  17. for (int i = 0 ; i< contests.size(); i++){
  18. imp = imp + contests[i][1];
  19. sum = sum + contests[i][0];
  20. vect[i] = contests[i][1];
  21. }
  22. //cout << imp << endl;
  23.  
  24. if (imp < k) {
  25. return sum;
  26. }
  27. bool cont = 1;
  28. while(cont){
  29.  
  30. idxs =
  31.  
  32.  
  33.  
  34. }
  35.  
  36.  
  37. return out;
  38.  
  39. }
  40.  
  41. int main()
  42. {
  43. ofstream fout(getenv("OUTPUT_PATH"));
  44.  
  45. string nk_temp;
  46. getline(cin, nk_temp);
  47.  
  48. vector<string> nk = split_string(nk_temp);
  49.  
  50. int n = stoi(nk[0]);
  51.  
  52. int k = stoi(nk[1]);
  53.  
  54. vector<vector<int>> contests(n);
  55. for (int i = 0; i < n; i++) {
  56. contests[i].resize(2);
  57.  
  58. for (int j = 0; j < 2; j++) {
  59. cin >> contests[i][j];
  60. }
  61.  
  62. cin.ignore(numeric_limits<streamsize>::max(), '\n');
  63. }
  64.  
  65. int result = luckBalance(k, contests);
  66.  
  67. fout << result << "\n";
  68.  
  69. fout.close();
  70.  
  71. return 0;
  72. }
  73.  
  74. vector<string> split_string(string input_string) {
  75. string::iterator new_end = unique(input_string.begin(), input_string.end(), [] (const char &x, const char &y) {
  76. return x == y and x == ' ';
  77. });
  78.  
  79. input_string.erase(new_end, input_string.end());
  80.  
  81. while (input_string[input_string.length() - 1] == ' ') {
  82. input_string.pop_back();
  83. }
  84.  
  85. vector<string> splits;
  86. char delimiter = ' ';
  87.  
  88. size_t i = 0;
  89. size_t pos = input_string.find(delimiter);
  90.  
  91. while (pos != string::npos) {
  92. splits.push_back(input_string.substr(i, pos - i));
  93.  
  94. i = pos + 1;
  95. pos = input_string.find(delimiter, i);
  96. }
  97.  
  98. splits.push_back(input_string.substr(i, min(pos, input_string.length()) - i + 1));
  99.  
  100. return splits;
  101. }
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