CodeForces 492D Vanya and Computer Game

D. Vanya and Computer Game

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Vanya and his friend Vova play a computer game where they need to destroy n monsters to pass a level. Vanya's character performs attack with frequency x hits per second and Vova's character performs attack with frequency y hits per second. Each character spends fixed time to raise a weapon and then he hits (the time to raise the weapon is 1 / x seconds for the first character and 1 / y seconds for the second one). The i-th monster dies after he receives ai 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 integers ai (1 ≤ ai ≤ 109) — the number of hits needed do destroy thei-th monster.

Output

Print n lines. In the i-th line print word "Vanya", if the last hit on the i-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 2
1
2
3
4

output

Vanya
Vova
Vanya
Both

input

2 1 1
1
2

output

Both
Both

Note

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

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

题目大意：n只怪兽有各自被攻击致死的次数，Vanya每秒攻击怪兽x次，Vova每秒攻击怪兽y次。求每只怪物被消灭时最后攻击它的是谁，如果同时攻击，输出Both。

解题思路：二分查找，时间作变量，以攻击次数为判断条件。可以理解为Vanya用y秒攻击一次，Vova用x秒攻击一次，Lcm(x，y)秒时同时攻击。mid秒时攻击了(mid/x+mid/y)次。

解题过程：wa点：注意用大数。

代码如下：

#include <cstdio>
#define LL long long
int main()
{
    LL n,x,y,t;
    LL high,low,mid,a;
    scanf("%lld%lld%lld",&n,&x,&y);
    for(int i=0;i<n;i++)
    {
        scanf("%lld",&t);
        high=1e15,low=0;
        while(low<=high)
        {
            mid=(high+low)/2;
            if(mid/x+mid/y<t)//mid时间里A的攻击次数+B的攻击次数与怪物被攻击致死的次数比较。
                low=mid+1;
            else
            {
                a=mid;	//记录下怪物刚好死亡的时间
                high=mid-1;
            }
        }
        if(a%x==0&&a%y==0)//a秒时A刚好击打，a秒时B刚好击打
             printf("Both\n");
        else if(a%y)   
            printf("Vova\n");
        else
            printf("Vanya\n");
    }
    return 0;
}