# Golden rectangles! Hackerearth solution

Problem:-

You have N rectangles. A rectangle is golden if the ratio of its sides is in between $\left[1.6,1.7\right]$, both inclusive. Your task is to find the number of golden rectangles.

Input format

• First line: Integer $N$ denoting the number of rectangles
• Each of the $N$ following lines: Two integers $W,H$ denoting the width and height of a rectangle

Output format

• Print the answer in a single line.

Constraints

$1\le N\le {10}^{5}$

$1\le W,H\le {10}^{9}$

Sample Input
5
10 1
165 100
180 100
170 100
160 100


Sample Output
3

Time Limit: 1
Memory Limit: 256
Source Limit:
Explanation

There are three golden rectangles: (165, 100), (170, 100), (160, 100).

$10$

Code:-

Solution 1 ( C language):-

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
bool isInRatio(double a, double b){
return (a/b>=1.6 && a/b<=1.7);
}
int main () {
char first[10] = {0};
fgets(first, 10, stdin);
int N = strtoll(first, NULL, 10);
int sum = 0;
for(int i = 0; i<N; i++){
char line[100]={0};
fgets(line, 100, stdin);
char *end;
int W = strtoll(line, &end, 10);
int H = strtoll(end, NULL, 10);
if(isInRatio(W, H)||isInRatio(H,W)){
sum++;
}
}
printf("%d\n", sum);
return 0;
}

Solution 2 ( C++ language):-

#include <bits/stdc++.h>
using namespace std;
#define ll long long int
#define ld long double
#define rep(i, k, n) for (long long int i = k; i < n; i++)
#define rept(i, k, n) for (auto i = k; i != n; ++i)
#define drep(i, k, n) for (long long int i = k; i >= n; i--)
#define endl '\n'
const int MAX_N = 1e5 + 5;
const ll MOD = 1e9 + 7;
const ll INF = 1e9;
const ld EPS = 1e-9;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0); cout.tie(0);

ll t;
cin>>t;
int count=0;
while(t-->0)
{
int w,h;
cin>>w>>h;

if(w>=1.6*h && w<=1.7*h)
{
count++;

}
else if(h>=1.6*w && h<=1.7*w)
{
count++;
}
// cout<<rat<<endl;
}
cout<<count<<endl;
}

Solution 3 (java language):-

import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
public class GoldenRectangle {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
}
{
din = new DataInputStream(
new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
}
{
byte[] buf = new byte[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
if (cnt != 0) {
break;
}
else {
continue;
}
}
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
while (c <= ' ') {
}
boolean neg = (c == '-');
if (neg)
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
while (c <= ' ')
boolean neg = (c == '-');
if (neg)
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
while (c <= ' ')
boolean neg = (c == '-');
if (neg)
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
BUFFER_SIZE);