Automated Teller Machine (ATM) is an electronic device that enables people to withdraw cash from their bank account. Every ATM has a limit for number of currency notes (say N), it can give at a time.
A bank wants to design an ATM for school students. The unique feature of this ATM would be that it would always give maximum number of currency notes possible, to make the students happy. Available denomination of currency notes in the ATM are 100, 200, 500, 1000
Constraints:
N<100
Input Format:
First Line provides an integer, N
Second Line provides an integer denoting the amount you want to withdraw (in multiples of 100)
Third Line provides an integer denoting the available currency note of Rs 100 in the ATM
Fourth Line provides an integer denoting the available currency note of Rs 200 in the ATM
Fifth Line provides an integer denoting the available currency note of Rs 500 in the ATM
Sixth Line provides an integer denoting the available currency note of Rs 1000 in the ATM
Output:
One line containing the maximum number of currency note possible for the desired withdrawal amount. Output should be 0 (zero) if transaction is not possible, for example if sufficient fund is not available in the ATM.
Time Limit:
1
Explanation:
Example 1:
Input:
10 1300 10 10 10 10
Output:
10
Explanation:
Here,
7 * 100 + 3 * 200 + 0 * 500 + 0 * 1000 hence maximum possible currency = 10.
Example 2:
Input:
5 1700 1 2 2 2
Output:
3
Explanation:
Here,
0 * 100 + 1 * 200 + 1 * 500 + 1 * 1000 hence maximum possible currency = 3.
n = int(input()) # Maximum number of currency notes allowed to be withdrawn.
amount = int(input()) # The desired withdrawal amount (in multiples of 100).
hundred = int(input()) # Available currency notes of Rs 100.
twohundred = int(input()) # Available currency notes of Rs 200.
fivehundred = int(input()) # Available currency notes of Rs 500.
thousand = int(input()) # Available currency notes of Rs 1000.
count = 0
for i in range(hundred + 1):
sum = i * 100
if sum == amount and i <= n and i > count:
count = i
if sum < amount:
for j in range(twohundred + 1):
sum1 = sum + j * 200
if sum1 == amount and (i + j) <= n and (j + i) > count:
count = i + j
if sum1 < amount:
for k in range(fivehundred + 1):
sum2 = sum1 + k * 500
if sum2 == amount and (i + j + k) <= n and (i + j + k) > count:
count = i + j + k
if sum2 < amount:
for l in range(thousand + 1):
sum3 = sum2 + l * 1000
if sum3 == amount and (i + j + k + l) <= n and (i + j + k + l) > count:
count = i + j + k + l
if sum3 > amount:
l = thousand + 1 # Exit the loop if sum3 exceeds the desired amount.
print(count) # Output the maximum possible currency note count.
#include <stdio.h>
int main() {
int sum, sum1, sum2, sum3, amount, n, hundred, twohundred, fivehundred, thousand, i, j, k, l, count = 0;
scanf("%d", &n); // Maximum number of currency notes allowed to be withdrawn.
scanf("%d", &amount); // The desired withdrawal amount (in multiples of 100).
scanf("%d", &hundred); // Available currency notes of Rs 100.
scanf("%d", &twohundred); // Available currency notes of Rs 200.
scanf("%d", &fivehundred); // Available currency notes of Rs 500.
scanf("%d", &thousand); // Available currency notes of Rs 1000.
for (i = 0; i <= hundred; i++) {
sum = i * 100;
if (sum == amount && i <= n && i > count)
count = i;
if (sum < amount)
for (j = 0; j <= twohundred; j++) {
sum1 = sum + j * 200;
if (sum1 == amount && (i + j) <= n && (j + i) > count)
count = i + j;
if (sum1 < amount)
for (k = 0; k <= fivehundred; k++) {
sum2 = sum1 + k * 500;
if (sum2 == amount && (i + j + k) <= n && (i + j + k) > count)
count = i + j + k;
if (sum2 < amount)
for (l = 0; l <= thousand; l++) {
sum3 = sum2 + l * 1000;
if (sum3 == amount && (i + j + k + l) <= n && (i + j + k + l) > count)
count = i + j + k + l;
if (sum3 > amount)
l = thousand + 1; // Exit the loop if sum3 exceeds the desired amount.
}
}
}
}
printf("%dn", count); // Output the maximum possible currency note count.
return 0;
}
#include<bits/stdc++.h>
using namespace std;
int main() {
int sum, sum1, sum2, sum3, amount, n, hundred, twohundred, fivehundred, thousand, i, j, k, l, count = 0;
cin >> n; // Maximum number of currency notes allowed to be withdrawn.
cin >> amount; // The desired withdrawal amount (in multiples of 100).
cin >> hundred; // Available currency notes of Rs 100.
cin >> twohundred; // Available currency notes of Rs 200.
cin >> fivehundred; // Available currency notes of Rs 500.
cin >> thousand; // Available currency notes of Rs 1000.
for (i = 0; i <= hundred; i++) {
sum = i * 100;
if (sum == amount && i <= n && i > count)
count = i;
if (sum < amount)
for (j = 0; j <= twohundred; j++) {
sum1 = sum + j * 200;
if (sum1 == amount && (i + j) <= n && (j + i) > count)
count = i + j;
if (sum1 < amount)
for (k = 0; k <= fivehundred; k++) {
sum2 = sum1 + k * 500;
if (sum2 == amount && (i + j + k) <= n && (i + j + k) > count)
count = i + j + k;
if (sum2 < amount)
for (l = 0; l <= thousand; l++) {
sum3 = sum2 + l * 1000;
if (sum3 == amount && (i + j + k + l) <= n && (i + j + k + l) > count)
count = i + j + k + l;
if (sum3 > amount)
l = thousand + 1; // Exit the loop if sum3 exceeds the desired amount.
}
}
}
}
cout << count << endl; // Output the maximum possible currency note count.
return 0;
}
using System;
class Program {
static void Main() {
int sum, sum1, sum2, sum3, amount, n, hundred, twohundred, fivehundred, thousand, i, j, k, l, count = 0;
n = int.Parse(Console.ReadLine()); // Maximum number of currency notes allowed to be withdrawn.
amount = int.Parse(Console.ReadLine()); // The desired withdrawal amount (in multiples of 100).
hundred = int.Parse(Console.ReadLine()); // Available currency notes of Rs 100.
twohundred = int.Parse(Console.ReadLine()); // Available currency notes of Rs 200.
fivehundred = int.Parse(Console.ReadLine()); // Available currency notes of Rs 500.
thousand = int.Parse(Console.ReadLine()); // Available currency notes of Rs 1000.
for (i = 0; i <= hundred; i++) {
sum = i * 100;
if (sum == amount && i <= n && i > count)
count = i;
if (sum < amount)
for (j = 0; j <= twohundred; j++) {
sum1 = sum + j * 200;
if (sum1 == amount && (i + j) <= n && (j + i) > count)
count = i + j;
if (sum1 < amount)
for (k = 0; k <= fivehundred; k++) {
sum2 = sum1 + k * 500;
if (sum2 == amount && (i + j + k) <= n && (i + j + k) > count)
count = i + j + k;
if (sum2 < amount)
for (l = 0; l <= thousand; l++) {
sum3 = sum2 + l * 1000;
if (sum3 == amount && (i + j + k + l) <= n && (i + j + k + l) > count)
count = i + j + k + l;
if (sum3 > amount)
l = thousand + 1; // Exit the loop if sum3 exceeds the desired amount.
}
}
}
}
Console.WriteLine(count); // Output the maximum possible currency note count.
}
}
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
int sum, sum1, sum2, sum3, amount, n, hundred, twohundred, fivehundred, thousand, i, j, k, l, count = 0;
Scanner scanner = new Scanner(System.in);
n = scanner.nextInt(); // Maximum number of currency notes allowed to be withdrawn.
amount = scanner.nextInt(); // The desired withdrawal amount (in multiples of 100).
hundred = scanner.nextInt(); // Available currency notes of Rs 100.
twohundred = scanner.nextInt(); // Available currency notes of Rs 200.
fivehundred = scanner.nextInt(); // Available currency notes of Rs 500.
thousand = scanner.nextInt(); // Available currency notes of Rs 1000.
for (i = 0; i <= hundred; i++) {
sum = i * 100;
if (sum == amount && i <= n && i > count)
count = i;
if (sum < amount)
for (j = 0; j <= twohundred; j++) {
sum1 = sum + j * 200;
if (sum1 == amount && (i + j) <= n && (j + i) > count)
count = i + j;
if (sum1 < amount)
for (k = 0; k <= fivehundred; k++) {
sum2 = sum1 + k * 500;
if (sum2 == amount && (i + j + k) <= n && (i + j + k) > count)
count = i + j + k;
if (sum2 < amount)
for (l = 0; l <= thousand; l++) {
sum3 = sum2 + l * 1000;
if (sum3 == amount && (i + j + k + l) <= n && (i + j + k + l) > count)
count = i + j + k + l;
if (sum3 > amount)
l = thousand + 1; // Exit the loop if sum3 exceeds the desired amount.
}
}
}
}
System.out.println(count); // Output the maximum possible currency note count.
}
}
