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A. Elections***the Country ofByalechinsk is running elections involving n candidates. The country consists ofM cities. We know how many peopleinch eachCity voted for eachCandidate. The electoralsystem inch theCountry is pretty unusual. At the FirstStage ofElections theVotes is counted for eachCityitis assumed thatinch eachCity won theCandidate who got theHighest Number ofVotesinchThis city, and ifSeveral candidates got theMaximum Number ofVotes, Then theWinner is the One with aSmaller index. At the SecondStage ofElections theWinner is determined by theSame principle over theCities theWinner of theElections is theCandidate who woninch theMaximum Number ofCities and amongThose who got theMaximum Number ofCities theWinner is the One with aSmaller index. Determine who'll win theElections. Inputthe First Line of theInputcontains Bothintegers n, m (1? ≤?n,?m?≤? -) — the Number ofCandidates and ofCities, respectively. each of theNext mLines containsn non-negative integers, theJ-th Number inch theI-th LineAIJ (1? ≤?j?≤?n,1? ≤?i?≤?m,0? ≤?aij?≤?109) denotes the Number ofVotes forCandidate JinchCity i.it are guaranteed that theTotal Number ofPeopleinchAll theCities does notExceed109.OutputprintaSingle Number— theIndex of theCandidate who won theElections. The candidates is indexed starting from One.
Test instructions
Given n rows m column, the number of votes on behalf of the first city of the players of J Data[i][j], the election is divided into two times, for the first time, each city to elect the city of the highest ticket player to become the winner, if the number of votes, the election of the smallest, and then the second round, Select the players who have the most victories in each city as the final winner, if the number is the same, the number is the smallest.
Directly on the code:
/*date:2015-8-26 Night Author:itakmotto: Today I want to go beyond yesterday's me, tomorrow I will surpass today's me, to create better code as the goal, constantly surpass oneself. */#include <iostream>#include <cstring>using namespace STD;intdata[ the][ the];intans[ the];intMain () {intM, N, J, I, Ind; while(Cin>>m>>n) {memset(ANS,0,sizeof(ans)); for(i=0; i<n; i++) {intMax =-999; for(j=0; j<m; J + +) {Cin>>data[i][j];if(Data[i][j] > Max) {Max = data[i][j]; IND = j; }} ans[ind]++; }intMax =-9999; for(i=0; i<m; i++)if(Ans[i] > Max) {Max = ans[i]; IND = i; }cout<<ind+1<<endl; }return 0;}/**INPUT3 2 3 2 1output2input3 410 1 2 7output1**/ 5
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Codeforces 570 A. Elections