// Count first-place votes for (int i = 0; i < voters; i++) { for (int j = 0; j < candidates; j++) { if (j == 0) { candidates_list[voters_prefs[i].preferences[j] - 1].votes++; } } } }
The implementation involves the following functions: #include <stdio.h> #include <stdlib.h> Cs50 Tideman Solution
3 3 1 2 3 1 3 2 2 1 3 This input represents an election with 3 voters and 3 candidates. The output of the program should be: // Count first-place votes for (int i =