**Jim and the Orders**

### Problem Statement :

Jim's Burgers has a line of hungry customers. Orders vary in the time it takes to prepare them. Determine the order the customers receive their orders. Start by numbering each of the customers from to , front of the line to the back. You will then be given an order number and a preparation time for each customer. The time of delivery is calculated as the sum of the order number and the preparation time. If two orders are delivered at the same time, assume they are delivered in ascending customer number order. For example, there are customers in line. They each receive an order number and a preparation time .: Customer 1 2 3 4 5 Order # 8 5 6 2 4 Prep time 3 6 2 3 3 Calculate: Serve time 11 11 8 5 7 We see that the orders are delivered to customers in the following order: Order by: Serve time 5 7 8 11 11 Customer 4 5 3 1 2 Function Description Complete the jimOrders function in the editor below. It should return an array of integers that represent the order that customers' orders are delivered. jimOrders has the following parameter(s): orders: a 2D integer array where each is in the form . Input Format The first line contains an integer , the number of customers. Each of the next lines contains two space-separated integers, an order number and prep time for . Output Format Print a single line of space-separated customer numbers (recall that customers are numbered from to ) that describes the sequence in which the customers receive their burgers. If two or more customers receive their burgers at the same time, print their numbers in ascending order.

### Solution :

` ````
Solution in C :
in C :
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
typedef long long int ll;
typedef struct list
{
ll t;
ll d;
ll pos
}L;
int compare(const void *a,const void *b)
{
L *e1 = (L *)a;
L *e2 = (L *)b;
return (e1->d+e1->t)-(e2->t+e2->d);
}
L val[1005];
int main()
{
ll n,i;
scanf("%lld",&n);
for(i=0;i<n;i++)
{
scanf("%lld%lld",&val[i].t,&val[i].d);
val[i].pos=i+1;
}
qsort(val,n,sizeof(L),compare);
for(i=0;i<n;i++)
{
printf("%lld ",val[i].pos);
}
printf("\n");
return 0;
}
```

` ````
Solution in C++ :
in C++ :
#include <bits/stdc++.h>
using namespace std;
int N;
pair<int, int> A[1000];
int main()
{
scanf("%d", &N);
int a, b;
for(int i=0; i<N; i++)
{
scanf("%d%d", &a, &b);
A[i]=make_pair(a+b, i+1);
}
sort(A, A+N);
for(int i=0; i<N; i++)
printf("%d ", A[i].second);
printf("\n");
return 0;
}
```

` ````
Solution in Java :
In Java :
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Comparator;
import java.util.InputMismatchException;
public class A {
static InputStream is;
static PrintWriter out;
static String INPUT = "";
static void solve()
{
int n = ni();
int[][] a = new int[n][];
for(int i = 0;i < n;i++){
a[i] = new int[]{ni() + ni(), i};
}
Arrays.sort(a, new Comparator<int[]>() {
public int compare(int[] a, int[] b) {
if(a[0] != b[0])return a[0] - b[0];
return a[1] - b[1];
}
});
for(int i = 0;i < n;i++){
if(i > 0)out.print(" ");
out.print(a[i][1]+1);
}
out.println();
}
public static void main(String[] args) throws Exception
{
long S = System.currentTimeMillis();
is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
out = new PrintWriter(System.out);
solve();
out.flush();
long G = System.currentTimeMillis();
tr(G-S+"ms");
}
private static boolean eof()
{
if(lenbuf == -1)return true;
int lptr = ptrbuf;
while(lptr < lenbuf)if(!isSpaceChar(inbuf[lptr++]))return false;
try {
is.mark(1000);
while(true){
int b = is.read();
if(b == -1){
is.reset();
return true;
}else if(!isSpaceChar(b)){
is.reset();
return false;
}
}
} catch (IOException e) {
return true;
}
}
private static byte[] inbuf = new byte[1024];
static int lenbuf = 0, ptrbuf = 0;
private static int readByte()
{
if(lenbuf == -1)throw new InputMismatchException();
if(ptrbuf >= lenbuf){
ptrbuf = 0;
try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
if(lenbuf <= 0)return -1;
}
return inbuf[ptrbuf++];
}
private static boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
private static int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
private static double nd() { return Double.parseDouble(ns()); }
private static char nc() { return (char)skip(); }
private static String ns()
{
int b = skip();
StringBuilder sb = new StringBuilder();
while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ')
sb.appendCodePoint(b);
b = readByte();
}
return sb.toString();
}
private static char[] ns(int n)
{
char[] buf = new char[n];
int b = skip(), p = 0;
while(p < n && !(isSpaceChar(b))){
buf[p++] = (char)b;
b = readByte();
}
return n == p ? buf : Arrays.copyOf(buf, p);
}
private static char[][] nm(int n, int m)
{
char[][] map = new char[n][];
for(int i = 0;i < n;i++)map[i] = ns(m);
return map;
}
private static int[] na(int n)
{
int[] a = new int[n];
for(int i = 0;i < n;i++)a[i] = ni();
return a;
}
private static int ni()
{
int num = 0, b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-'){
minus = true;
b = readByte();
}
while(true){
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
}else{
return minus ? -num : num;
}
b = readByte();
}
}
private static long nl()
{
long num = 0;
int b;
boolean minus = false;
while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
if(b == '-'){
minus = true;
b = readByte();
}
while(true){
if(b >= '0' && b <= '9'){
num = num * 10 + (b - '0');
}else{
return minus ? -num : num;
}
b = readByte();
}
}
private static void tr(Object... o) { if(INPUT.length() != 0)System.out.println(Arrays.deepToString(o)); }
}
```

` ````
Solution in Python :
tuples = []
a = int(input())
for i in range(a):
x, y = map(int, input().split(' '))
tuples.append((x+y, i+1))
tuples.sort(key = lambda x:x[0])
print(' '.join([str(x[1]) for x in tuples]))
```

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