codeforces 492D. 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 the i-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 time 1 / 2, Vanya makes the third hit at time2 / 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 time 1.

这题思路不难，我的思路是先找出最后一秒t，然后m减去(t-1)*(x+y),然后二分最后一秒。但是这题非常卡精度，非常坑啊。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
#define inf 0x7fffffff
#define eps 1e-9
#define eps1 1e-6

double x,y,m;
int cal1(double shijian)
{
    int i,j,num1,num2;
    int l=0;
    int r=x,mid;
    while(l<=r){
        mid=(l+r)/2;
        if(1.0*mid*(1.0/(x*1.0) )>shijian )r=mid-1;
        else l=mid+1;
    }
    return r;
}


int cal2(double shijian)
{
    int i,j,num1,num2;
    int l=0;
    int r=y,mid;
    while(l<=r){
        mid=(l+r)/2;
        if(1.0*mid*(1.0/ (1.0*y) )>shijian )r=mid-1;
        else l=mid+1;
    }
    return r;

}

int main()
{
    int n,i,j,t1,num11,num22,h;
    double l,r,mid;
    int num1,num2;
    while(scanf("%d%lf%lf",&n,&x,&y)!=EOF)
    {
        for(h=1;h<=n;h++){
            scanf("%lf",&m);
            t1=m/(x+y);
            m-=(x+y)*t1;
            l=0;r=1;
            for(i=1;i<=200;i++){
                mid=(l+r)/2.0;
                num1=cal1(mid);
                num2=cal2(mid);
                if(num1+num2>=m)r=mid;
                //if(floor(mid*x)+floor(mid*y)>=m)r=mid;
                else l=mid;
            }
            double num3=floor(mid*x+eps);
            double num4=floor(mid*y+eps);  //这里不加eps会错的= =
            if(num3*y==num4*x)printf("Both
");
            else if(num3*y>num4*x)printf("Vanya
");
            else if(num3*y<num4*x)printf("Vova
");
        }
    }
    return 0;
}