# Ice Cream Parlor

### Problem Statement :

```Two friends like to pool their money and go to the ice cream parlor. They always choose two distinct flavors and they spend all of their money.

Given a list of prices for the flavors of ice cream, select the two that will cost all of the money they have.

Example.

The two flavors that cost  and  meet the criteria. Using -based indexing, they are at indices  and .

Function Description

Complete the icecreamParlor function in the editor below.

icecreamParlor has the following parameter(s):

int m: the amount of money they have to spend
int cost[n]: the cost of each flavor of ice cream
Returns

int[2]: the indices of the prices of the two flavors they buy, sorted ascending

Input Format

The first line contains an integer, , the number of trips to the ice cream parlor. The next  sets of lines each describe a visit.

Each trip is described as follows:

The integer , the amount of money they have pooled.
The integer , the number of flavors offered at the time.
space-separated integers denoting the cost of each flavor: .
Note: The index within the cost array represents the flavor of the ice cream purchased.```

### Solution :

```                            ```Solution in C :

In   C++  :

#include <list>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
#include <numeric>
#include <cstdlib>
#include <cmath>
#include <cstdio>
#include <fstream>
#include <ctime>

using namespace std; // }}}

int main()
{
//time_t start, end;
//time(&start);

//ifstream cin("test.in");
//ofstream cout("test.out");
//fcout.precision(6);
//fcout.setf(ios::fixed,ios::floatfield);

int T;
cin >> T;
for (int i = 0; i < T; ++i) {
int C, L;
cin >> C >> L;
vector<int> v(L);
map<int, int> M;
for (int j = 0; j < L; ++j)
cin >> v[j];

for (int j = 0; j < L; ++j) {
int diff = C - v[j];
if (M.count(diff)) {
cout << 1 + M[diff] << " " << 1 + j << endl;
break;
}
M[v[j]] = j;
}
}

//time(&end);
//cout << difftime(end, start) << endl;

return 0;
}

In  Java  :

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

public static void main(String[] args) {
Scanner sc = new Scanner(System.in);

int n = sc.nextInt();
for(int k = 0; k < n; k++){
int C = sc.nextInt();
int P = sc.nextInt();
int [] p = new int[P+1];

for(int i = 1; i <= P;i++)p[i] = sc.nextInt();

for(int i = 1; i <= P; i++){
for(int j = i+1; j <= P; j++){
if(p[i]+p[j] == C){
System.out.println(i + " " + j);
break;
}
}
}
}
sc.close();
}
}

In   C  :

#include<stdio.h>
int main()
{
int t,c,l,i,j,arr[20000];
for(scanf("%d",&t);t>0;t--)
{
scanf("%d%d",&c,&l);
for(i=0;i<l;i++)
scanf("%d",&arr[i]);
for(i=0;i<l-1;i++)
for(j=i+1;j<l;j++)
{
if(arr[i]+arr[j]==c)
printf("%d %d\n",i+1,j+1);
}
}
return 0;
}

In   Python3 :

def main():
# Loop through for each test.
for _ in range(int(input())):
dollars = int(input())
numFlavors = int(input())
flavors = [int(i) for i in input().split()]

# Determine the correct indexes.
index1, index2 = -1, -1
for i in range(numFlavors):
for j in range(numFlavors):
if (i != j) and (flavors[i] + flavors[j] == dollars):
index1 = i + 1
index2 = j + 1
break
if index1 != -1 and index2 != -1:
break

print(str(index1) + " " + str(index2))

if __name__ == "__main__":
main()```
```

## Truck Tour

Suppose there is a circle. There are N petrol pumps on that circle. Petrol pumps are numbered 0 to (N-1) (both inclusive). You have two pieces of information corresponding to each of the petrol pump: (1) the amount of petrol that particular petrol pump will give, and (2) the distance from that petrol pump to the next petrol pump. Initially, you have a tank of infinite capacity carrying no petr

## Queries with Fixed Length

Consider an -integer sequence, . We perform a query on by using an integer, , to calculate the result of the following expression: In other words, if we let , then you need to calculate . Given and queries, return a list of answers to each query. Example The first query uses all of the subarrays of length : . The maxima of the subarrays are . The minimum of these is . The secon

## QHEAP1

This question is designed to help you get a better understanding of basic heap operations. You will be given queries of types: " 1 v " - Add an element to the heap. " 2 v " - Delete the element from the heap. "3" - Print the minimum of all the elements in the heap. NOTE: It is guaranteed that the element to be deleted will be there in the heap. Also, at any instant, only distinct element