Algorithm
Problem Name: 2 AD-HOC - beecrowd | 2469
Problem Link: https://www.beecrowd.com.br/judge/en/problems/view/2469
Grades
By OBI - Olimpíada Brasileira de Informática 2014 
Brazil
Timelimit: 1
Professor Archimedes needs your help to find out which is the most common notes between the notes that students took in his last race. The class has N students and your program must print the note that appears more often in N notes list. If more than more frequent note, you must print the biggest one! For example, if the class has N = 10 students and grades are [20, 25, 85, 40, 25, 90, 25, 40, 55, 40], the most common grades are 25 and 40, taking place three times each . Your program should then print 40.
Input
The input consists of two lines. The first line contains an integer N, the number of students in the class. The second line contains N integers, which is the list of students' grades.
Output
Your program must print a single line containing just a number, the most frequent note of the list.
| Input Sample | Output Sample |
|
10 |
40 |
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define true 1
#define false 0
#define MAXSIZE 100100
typedef struct Notas{
unsigned nota;
unsigned posicao;
} Notas;
int compara(Notas *, Notas *);
int main (void)
{
unsigned i, n, tmp;
Notas notas[MAXSIZE];
scanf("%u", &n);
memset(notas, 0, sizeof(notas));
while (n--)
{
scanf("%u", &tmp);
notas[tmp].nota++;
notas[tmp].posicao = tmp;
}
qsort(notas, MAXSIZE, sizeof(Notas), compara);
printf("%u\n", notas[0].posicao);
}
int compara(Notas *a, Notas *b)
{
if (a->nota == b->nota)
{
if (a->posicao > b->posicao)
return -1;
else
return 1;
}
else if (a->nota > b->nota)
return -1;
else
return 1;
}
Input
20 25 85 40 25 90 25 40 55 40
Output
#2 Code Example with C++ Programming
Code -
C++ Programming
#include <cstdio>
#include <map>
using namespace std;
int main() {
int n, i, resposta, vezes = 0, davez;
map < int, int> mapa;
map<int, int>::iterator it;
scanf("%d", &n);
for (i = 0; i < n; i++) {
scanf("%d", &davez);
if (mapa.find(davez) != mapa.end())
mapa[davez] += 1;
else
mapa[davez] = 1;
}
for (it = mapa.begin(); it != mapa.end(); it++) {
if ((*it).second > vezes) {
resposta = (*it).first;
vezes = (*it).second;
} else if ((*it).second == vezes && (*it).first > resposta)
resposta = (*it).first;
}
printf("%d\n", resposta);
return 0;
}
Input
20 25 85 40 25 90 25 40 55 40
Output