Algorithm

B. Color the Fence
time limit per test
2 seconds
memory limit per test
256 megabytes
Problem link- https://codeforces.com/problemset/problem/349/B
input
standard input
output
standard output

Igor has fallen in love with Tanya. Now Igor wants to show his feelings and write a number on the fence opposite to Tanya's house. Igor thinks that the larger the number is, the more chance to win Tanya's heart he has.

Unfortunately, Igor could only get v liters of paint. He did the math and concluded that digit d requires ad liters of paint. Besides, Igor heard that Tanya doesn't like zeroes. That's why Igor won't use them in his number.

Help Igor find the maximum number he can write on the fence.

Input

The first line contains a positive integer v (0 ≤ v ≤ 106). The second line contains nine positive integers a1, a2, ..., a9 (1 ≤ ai ≤ 105).

Output

Print the maximum number Igor can write on the fence. If he has too little paint for any digit (so, he cannot write anything), print -1.

Examples
input
Copy
5
5 4 3 2 1 2 3 4 5
output
Copy
55555
input
Copy
2
9 11 1 12 5 8 9 10 6
output
Copy
33
input
Copy
0
1 1 1 1 1 1 1 1 1
output
Copy
-1



 

Code Examples

#1 Code Example with C++ Programming

Code - C++ Programming

#include <stdio.h>

using namespace std;

int arr[9];

int main() {
	int n;
	scanf("%i", &n);
	for (int i = 0; i < 9; i++)
	    scanf("%i", &arr[i]);

	bool f = true;
	for (int i = 0; i < 9; i++)
		if (n >= arr[i])
			f = false;

	if (f)
		puts("-1");
	else {
		int min_v = 1e9, min_n = 1e9, mx_v, mx_n;
		for (int i = 0; i < 9; i++)
			if (arr[i] <= min_v) {
				min_v = arr[i];
				min_n = i;
			}

		int digits = n / min_v;
		int esc = n % min_v;

		if (!esc) {
			for (int i = 0; i < digits; i++)
				printf("%i", min_n + 1);
			putchar('\n');
		} else {
			while (esc) {
				digits--;
				esc += arr[min_n];

				mx_v = 1e9, mx_n = 1e9;

				for (int i = 0; i < 9; i++)
					if (arr[i] <= esc) {
						mx_v = arr[i];
						mx_n = i;
					}

				if ((mx_v == 1e9 && mx_n == 1e9) || mx_n == min_n) {
					digits++;
					esc -= arr[min_n];
					break;
				}

				esc -= mx_v;
				printf("%i", mx_n + 1);
			}

			for (int i = 0; i < digits; i++)
				printf("%i", min_n + 1);
			putchar('\n');
		}
	}

	return 0;
}
Copy The Code & Try With Live Editor

Input

x
–
+
cmd
5
5 4 3 2 1 2 3 4 5

Output

x
–
+
cmd
55555
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