Algorithm

C. Vanya and Scales
time limit per test
1 second
memory limit per test
256 megabytes
Problem link- https://codeforces.com/contest/552/problem/C
input
standard input
output
standard output

Vanya has a scales for weighing loads and weights of masses w0, w1, w2, ..., w100 grams where w is some integer not less than 2 (exactly one weight of each nominal value). Vanya wonders whether he can weight an item with mass m using the given weights, if the weights can be put on both pans of the scales. Formally speaking, your task is to determine whether it is possible to place an item of mass m and some weights on the left pan of the scales, and some weights on the right pan of the scales so that the pans of the scales were in balance.

Input

The first line contains two integers w, m (2 ≤ w ≤ 109, 1 ≤ m ≤ 109) — the number defining the masses of the weights and the mass of the item.

Output

Print word 'YES' if the item can be weighted and 'NO' if it cannot.

Examples
input
Copy
3 7
output
Copy
YES
input
Copy
100 99
output
Copy
YES
input
Copy
100 50
output
Copy
NO
Note

Note to the first sample test. One pan can have an item of mass 7 and a weight of mass 3, and the second pan can have two weights of masses 9 and 1, correspondingly. Then 7 + 3 = 9 + 1.

Note to the second sample test. One pan of the scales can have an item of mass 99 and the weight of mass 1, and the second pan can have the weight of mass 100.

Note to the third sample test. It is impossible to measure the weight of the item in the manner described in the input.



 

Code Examples

#1 Code Example with C++ Programming

Code - C++ Programming

#include <stdio.h>
#include <vector>
#include <algorithm>

using namespace std;

int w, m;
vector<long long> powers, all[2];

void rec(int i, long long sum, int end, int idx) {
	if(i == end) {
		all[idx].push_back(sum);
		return;
	}

	rec(i + 1, sum, end, idx);
	rec(i + 1, sum + powers[i], end, idx);
	rec(i + 1, sum - powers[i], end, idx);
}

int main() {
	scanf("%d %d", &w, &m);

	if(w == 2) {
		puts("YES");
		return 0;
	}

	long long cur = 1, sum = 0;
	do {
		powers.push_back(cur);
		sum += cur;
		cur *= w;
	} while (cur - sum <= m);

	rec(0, 0, powers.size() / 2, 0);
	rec(powers.size() / 2, 0, powers.size(), 1);

	sort(all[0].begin(), all[0].end());

	for(int i = 0; i < (int)all[1].size(); ++i)
		if(binary_search(all[0].begin(), all[0].end(), m + all[1][i])) {
			puts("YES");
			return 0;
		}

	puts("NO");

	return 0;
}
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Input

x
–
+
cmd
3 7

Output

x
–
+
cmd
YES
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Demonstration

Codeforces Solution-Vanya and Scales-Solution in C, C++, Java, Python

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