## Algorithm

Problem Name: beecrowd | 1983

# The Chosen

By Victor Jatobá, UNIME Brazil

Timelimit: 1

Everyone is talking about the Jatoba Teacher classes. The MEC representatives came to the UNIME of Lauro de Freitas to know more details about this new way of teaching algorithms. In addition, they wanted to select one student to participate in the OBI-Tec (Olympiad Informatics Technical Level) and represent Kroton in the competition, because they know that there are the best. To select the best, they have available a list of the registration number of each student and their respective notes in the discipline. Your task is to help the Ministry of Education staff to find students better able to represent the institution and who knows guarantee your place. If the highest score is not greater than or equal to 8, you should print "Minimum note not reached" .

## Input

The file first contains the number of students (3 <= n <= 100) and then the n students bearing the registration number (0 < m < 1000000) each, followed by the note (0 <= note <= 10.0, with 1 decimal place).

Obs .: the notes will not be repeated. In other words, has no chance to have two students with the same note.

## Output

You must print the student's registration number with the highest score or "Minimum note not reached" (without quotes) if no student has taken greater note than or equal to 8.

 Input Sample Output Sample 3 1000 5 1001 10 1002 6 1001

 4 900775 5.7 201553 7.9 5032 6.2 2088 2.1 Minimum note not reached

## Code Examples

### #1 Code Example with C Programming

```Code - C Programming```

``````
#include<stdio.h>

int main()
{   int n,i,a,c = 0,ind[1000],p;
float gr[1000],b,max;

scanf("%d",&n);

for(i = 0;i  <  n; i++){
scanf("%d %f",&a,&b);
ind[i] = a;
gr[i] = b;
if(gr[i] >= 8){
if(max < gr[i]){
max=gr[i];
p = i;
//printf("%.2f \t %.2f",max,gr[i]);
}
c=1;
}
}
if(c == 1){
printf("%d\n",ind[p]);
}
else{
printf("Minimum note not reached\n">;
}

return 0;
}
``````
Copy The Code &

Input

cmd
3 1000 5 1001 10 1002 6

Output

cmd
1001

### #2 Code Example with C++ Programming

```Code - C++ Programming```

``````
#include<stdio.h>
int main()
{
int test;
scanf("%d",&test);
int code[test],i,final_value;
double frac[test],val=0.0;
for(i = 0; i  <  test; i++)
{
scanf("%d %lf",&code[i],&frac[i]);
}
for(i = 0; i  <  test; i++)
{
if(frac[i] > val)
{
val=frac[i];
final_value=i;
}
}
if(val<8)
printf("Minimum note not reached\n");
else
printf("%d\n",code[final_value]>;
return 0;
}
``````
Copy The Code &

Input

cmd
3 1000 5 1001 10 1002 6

Output

cmd
1001

### #3 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
var lines = input.split('\n');

const N = lines.shift();
let [a , b] = lines.shift().split(" ");
let tempMatricula;
let tempNota;
let i = 0;

while(i  <  N){
if(i == 0){
tempMatricula = a;
tempNota = b;
[a , b] = lines.shift().split(" ");
i++;
}
else{
if(parseFloat(tempNota) > parseFloat(b)){
i++;
[a , b] = lines.shift().split(" ");
}
else{
tempMatricula = a;
tempNota = b;
i++;
[a , b] = lines.shift().split(" ");
}
}
}

if(tempNota >= 8){
console.log(tempMatricula);
}
else{
console.log("Minimum note not reached");
}
``````
Copy The Code &

Input

cmd
3 1000 5 1001 10 1002 6

Output

cmd
1001

### #4 Code Example with Python Programming

```Code - Python Programming```

``````
n = int(input())
m = 0.0
al = 0
while(n > 0):
n -= 1
a, b = input().split()
if float(b) > m:
m = float(b)
al = int(a)

if m >= 8:
print(al)
else:
print("Minimum note not reached")
``````
Copy The Code &

Input

cmd
4 900775 5.7 201553 7.9 5032 6.2 2088 2.1

Output

cmd
Minimum note not reached