Codeforces 492D Vanya and Computer Game

最新推荐文章于 2017-10-26 17:45:06 发布

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最新推荐文章于 2017-10-26 17:45:06 发布

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原文链接：http://www.cnblogs.com/hlmark/p/4143634.html

本文介绍了一个关于Vanya和他的朋友Vova在一个电脑游戏中如何共同攻击怪物的问题，并提出了一种有效的算法解决方案。通过预处理(x+y)枪击轮次并进行排序，可以快速判断每个怪物的最后一击是由Vanya还是Vova完成，或者两者同时完成。

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

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.

可以知道。只需要算出a%(x+y)是谁打的就得到了正确答案了。

因为前面的时间两个人打得枪数都是固定了的。

那么预处理（x+y）枪（用整数处理就好了，直接用double的话有浮点误差），

标记一下是谁射的，然后进行排序。

O（1）看一下 a%(x+y) 是谁射的，看一下前后有没有相同时间不同人射的。

然后就可以弄出答案来了。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <algorithm>

using namespace std;
typedef long long LL;
typedef pair<LL,int> pii;
const int N = 2000011 ;
const int M = 2010 ;
const int inf = 1e9+7;
#define X first
#define Y second
LL n , x , y , a ;
vector<pii>e;
void Run() {
    e.clear();
    for( int i = 1 ; i <= x ; ++i ) e.push_back( pii( y*i ,0 ) );
    for( int i = 1 ; i <= y ; ++i ) e.push_back( pii( x*i ,1 ) );
    sort( e.begin() , e.end() ) ;
    for( int i = 0 ; i < n ; ++i ) {
        cin >> a ;
        LL s = a / ( x + y ) ;
        if( s * ( x + y ) < a ) a -= s * ( x + y ) ;
        else a = x + y ;
        a--;
        if( a < x + y && e[a].X == e[a+1].X ) { cout << "Both" << endl ;  continue ; }
        if( a > 0 && e[a].X == e[a-1].X ) { cout << "Both" << endl ;  continue ; }
        if( e[a].Y == 0 ) { cout << "Vanya"<<endl; continue ; }
        cout << "Vova" <<endl;
    }
}
int main()
{
    #ifdef LOCAL
        freopen("in.txt","r",stdin);
    #endif // LOCAL
    ios::sync_with_stdio(false);
    while( cin >> n >> x >> y ) Run();
}