codeforces 492D Vanya and Computer Game (二分)

D. Vanya and Computer Game

time limit per test

2 seconds

memory limit per test

256 megabytes

Vanya and his friend Vova play a computer game where they need to destroyn monsters to pass a level. Vanya's character performs attack with frequencyx hits per second and Vova's character performs attack with frequencyy hits per second. Each character spends fixed time to raise a weapon and then he hits (the time to raise the weapon is1 / x seconds for the first character and 1 / y seconds for the second one). The i-th monster dies after he receivesa_i hits.

Vanya and Vova wonder who makes the last hit on each monster. If Vanya and Vova make the last hit at the same time, we assume that both of them have made the last hit.

Input

The first line contains three integers n,x,y (1 ≤ n ≤ 10⁵,1 ≤ x, y ≤ 10⁶) — the number of monsters, the frequency of Vanya's and Vova's attack, correspondingly.

Next n lines contain integersa_i (1 ≤ a_i ≤ 10⁹) — the number of hits needed do destroy thei-th monster.

Output

Print n lines. In thei-th line print word "Vanya", if the last hit on thei-th monster was performed by Vanya, "Vova", if Vova performed the last hit, or "Both", if both boys performed it at the same time.

Sample test(s)

Input

4 3 21234

Output

VanyaVovaVanyaBoth

Input

2 1 112

Output

BothBoth

Note

In the first sample Vanya makes the first hit at time1 / 3, Vova makes the second hit at time 1 / 2, Vanya makes the third hit at time 2 / 3, and both boys make the fourth and fifth hit simultaneously at the time1.

In the second sample Vanya and Vova make the first and second hit simultaneously at time1.

题目链接：http://codeforces.com/problemset/problem/492/D

题目大意：两个人打怪，Vanya每秒打x次，每次打需要花费1/x秒修复武器，Vova每秒打y次，每次打需要1/y秒修复武器，一共有n只怪，第i只怪需要被打ai次，问第i只怪最后是被谁干掉的

题目分析：把1/xy当作一个单位时间，在t个单位时间内Vanya攻击了t/y次，Vova攻击了t/x次，所以我们只需要二分出杀死第a[i]个怪物要多少个单位时间all，显然最后的答案可以根据all对x和y的整除性判断出来，如果all对x整除等于0说明Vanya攻击的次数是整数次，而Vova不是，所以最后一次是由Vova攻击的，攻击完的总攻击值肯定大于杀死这个怪物所需要的值，另外两种情况同理

#include <iostream>#define ll long longusing namespace std;int main(){    ll n, x, y, hit, all;    cin >> n >> x >> y;    for(int i = 0; i < n; i++)    {        cin >> hit;        ll r = 1e18, l = 0, mid;        while(l <= r)        {            mid = (l + r) / 2;            if(mid / x + mid / y >= hit)            {                all = mid;                r = mid - 1;            }            else                l = mid + 1;        }        if(all % x)            cout << "Vanya" << endl;        else if(all % y)            cout << "Vova" << endl;        else            cout << "Both" << endl;    }}