In a normal election, one expects the percentages received by each of the candidates to sum to exactly 100 percent. There are two ways this might not be the case: if the election is fraudulent, or if the reported percentages are rounded. In a recent election, the number of voters was known to be exactly 10000. Assuming the election was run fairly, each voter voted for exactly one candidate. The percentage of the vote received by each candidate was rounded to the nearest whole number before being reported. Percentages lying halfway between two consecutive whole numbers were rounded up. The ministry of voting is looking for a proof of election fraud. You'll be given a int[] percentages, giving the reported percentage of the vote that went to each candidate. Return a String containing "YES" if the election is definitely fraudulent, otherwise return "NO" (quotes for clarity only). That is, return "YES" if it could not be the case that each of the 10000 voters voted for exactly one candidate, and "NO" otherwise.   Definition      Class: ElectionFraudDiv2 Method: IsFraudulent Parameters: int[] Returns: String Method signature: String IsFraudulent(int[] percentages) (be sure your method is public)        Constraints - percentages will contain between 1 and 50 elements, inclusive. - Each element of percentages will be between 0 and 100, inclusive.   Examples 0)      {100} Returns: "NO" If there's only one candidate, that candidate will receive 100% of the votes in a fair election. 1)      {34, 34, 34} Returns: "YES" Even accounting for rounding, these numbers are too high. 2)      {12, 12, 12, 12, 12, 12, 12, 12} Returns: "YES" These numbers are too low. 3)      {13, 13, 13, 13, 13, 13, 13, 13} Returns: "NO" Each candidate could have received exactly 1250 votes. 4)      {0, 1, 100} Returns: "NO" The only valid possibility is that the candidates received 0, 50, and 9950 votes, respectively. 5)      {3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 8, 9, 7, 9, 3, 2, 3, 8} Returns: "NO" You are given a list of of integers {0,5,100} that represent the percentages received by each of the candidates of an election. You are ask to write a program that detect if the election is fraudulent, or if the reported percentages are rounded. In that case, return "YES", otherwise, "NO". For more details of the problem statement please check the link below. In my opinion this problem was little bit more tricky than a normal DIV 2 250 the 19.10% of success rate confirm my thoughts... To solve this problem we need to consider the following three cases: Case #1: The sum of percentages equal to 100% In this case we assume the elections were not fraudulent. The answer for this case is "NO". Case #2: The sum of the percentages is greater than 100% In this case we need to check if the sum is greater than 100% because of a rounded up error. For example if the percentage for a certain candidate is 13% means that the real percentage value (not rounded) of the elections for that candidate is in the range [12.5%,13.49%]. To check that occurred a round error we substitute each of the candidate percentage for the minimum value in this range and sum them up. If we get a percentage sum less or equal to 100% means that the results are not fraudulent. The reason is because we can always get 100% with some combination of the rounded up error. In this case we should be careful with 0% is not possible to get 0% as a result of a rounded up error. For example: 3 candidates  A1={34%,34%,34%} sum_percentage=102 A2=33.5%,33.5%,33.5% sum_percentage=100.5 The results of this elections still fraudulent even after fixed the rounded up errors. Case #3: The sum of the percentages is less than 100% Similar to case #2 but taking the maximum value of the candidate percentage range. ? 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 class ElectionFraudDiv2 { public:    string IsFraudulent( vector <int> p ) {       int sum, N;       sum = N = 0;       for(int i = 0; i < sz(p); ++i) {          sum += p[i]*100;          if(p[i] > 0) ++N;       }       if(sum > 10000) {          if(sum - (N * 50) > 10000) return "YES";          else return "NO";       } else if(sum < 10000) {          if(sum + (sz(p) * 49) >= 10000) return "NO";          else return "YES";       }       return "NO";    } };