Codeforces Round #280 (Div. 2) D. Vanya and Computer Game 预处理

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 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 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 time 1.


#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
int dp[10000200];
int main()
{
    memset(dp,0,sizeof(dp));
    ll n,x,y;
    cin>>n>>x>>y;
    int flag2=0;
    int path1=1;
    int path2=1;
    int path=0;
    while(1)
    {
        if(path1==x&&path2==y)
            break;
        if(path1*y<path2*x)
        {
            dp[++path]=1;
            path1++;
        }
        else if(path1*y==path2*x)
        {
            dp[++path]=3;
            dp[++path]=3;
            path1++;
            path2++;
        }
        else
        {
            dp[++path]=2;
            path2++;
        }
        //cout<<dp[path]<<endl;
    }
    while(n--)
    {
        ll shengming;
        scanf("%I64d",&shengming);
        shengming=shengming%(x+y);
        if(shengming==x+y-1||shengming==0||dp[shengming]==3)
            printf("Both\n");
        else if(dp[shengming]==1)
        {
            if(flag2==0)
                printf("Vanya\n");
            else
                printf("Vova\n");
        }
        else
            if(flag2==0)
                printf("Vova\n");
            else
                printf("Vanya\n");
    }
    return 0;
}

 

posted @ 2014-12-02 09:13  qscqesze  阅读(306)  评论(0编辑  收藏  举报