http://codeforces.com/contest/492/problem/D

有时候感觉人sb还是sb,为什么题目都看不清楚?

x per second, y per second...

于是我们二分second即可。

而且对每个monster都有伤害。。。。

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << (#x) << " = " << (x) << endl
#define error(x) (!(x)?puts("error"):0)
#define rdm(x, i) for(int i=ihead[x]; i; i=e[i].next)
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; } int main() {
int n; ll x, y;
read(n); read(x); read(y);
for1(i, 1, n) {
ll a=getint();
ll l=0, r=1ll<<60, mid;
while(l<=r) {
mid=(l+r)>>1;
if(mid/x+mid/y>=a) r=mid-1;
else l=mid+1;
}
mid=r+1;
if(mid%x==0 && mid%y==0) puts("Both");
else if(mid%y==0) puts("Vanya");
else if(mid%x==0) puts("Vova");
}
return 0;
}

  


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 frequencyy 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.

【cf492】D. Vanya and Computer Game(二分)的更多相关文章

  1. Codeforces Round #280 (Div. 2) D. Vanya and Computer Game 二分

    D. Vanya and Computer Game Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contes ...

  2. cf492D Vanya and Computer Game

    D. Vanya and Computer Game time limit per test 2 seconds memory limit per test 256 megabytes input s ...

  3. 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 s ...

  4. Codeforces Round #280 (Div. 2) D. Vanya and Computer Game 数学

    D. Vanya and Computer Game time limit per test 2 seconds memory limit per test 256 megabytes input s ...

  5. CodeForces 492D Vanya and Computer Game (思维题)

    D. Vanya and Computer Game time limit per test 2 seconds memory limit per test 256 megabytes input s ...

  6. Codeforces 492D Vanya and Computer Game

    D. Vanya and Computer Game time limit per test 2 seconds memory limit per test 256 megabytes input s ...

  7. 数论 - Vanya and Computer Game

    Vanya and his friend Vova play a computer game where they need to destroy n monsters to pass a level ...

  8. 【Codeforces 492D】Vanya and Computer Game

    [链接] 我是链接,点我呀:) [题意] 题意 [题解] 第一个人攻击一次需要1/x秒 第二个人攻击一次需要1/y秒 这两个数字显然都是小数. 我们可以二分最后用了多少时间来攻击. 显然这个是有单调性 ...

  9. CodeForces Round #280 (Div.2)

    A. Vanya and Cubes 题意: 给你n个小方块,现在要搭一个金字塔,金字塔的第i层需要 个小方块,问这n个方块最多搭几层金字塔. 分析: 根据求和公式,有,按照规律直接加就行,直到超过n ...

随机推荐

  1. JAVA实现zip压缩需要注意的问题

    近来对院社二维码平台进行2.0升级改造.于昨日踩到一个巨坑.特此记录... 需求源于院社编辑在批量下载二维码的时候,系统后台需要对所要下载的二维码进行重命名和zip打包压缩. 系统测试的时候发现:首次 ...

  2. java反射--方法反射的基本操作

    方法的反射 1)如何获取某个方法 方法的名称和方法的参数列表才能唯一决定某个方法. 2)方法反射的操作 method.invoke(对象,参数列表). 代码实例: package com.reflec ...

  3. java面试题(开发框架)

    博客分类: java基础 面试Java多线程编程设计模式          java基础面试题目,以备不时之需 俗话说 细节决定成败.      就算很简单,很小的问题,我们还是要注意一下的.     ...

  4. 【centos6.5】安装LNMP(linux公社)

    1:查看环境: 1 2 [root@10-4-14-168 html]# cat /etc/redhat-release CentOS release 6.5 (Final) 2:关掉防火墙 1 [r ...

  5. Android 屏幕自适应方向尺寸

    最近感觉要被屏幕适配玩死了…… 安卓的手机为虾米不能像苹果那样只有几个分辨率呢?为什么呢!!!!!!!阿门…… 目前想到有两种解决办法…… 第一种:   HTML5+CSS3+WebView交互……目 ...

  6. Linux-软件包管理-yum在线管理-网络yum源

    cd /etc/yum.repos.d/  切换到etc目录下面的yum.repos.d这个目录中ls   查看当前linux系统的yum源文件信息,其中CentOS-Base.repo文件为默认的y ...

  7. 用ping让对方电脑堵塞瘫痪

    用ping让对方电脑堵塞瘫痪2008-04-27 11:32 定义echo数据包大小. 在默认的情况下windows的ping发送的数据包大小为32byt,我们也可以自己定义它的大小, 但有一个大小的 ...

  8. Lintcode---克隆二叉树

    深度复制一个二叉树. 给定一个二叉树,返回一个他的 克隆品 . 您在真实的面试中是否遇到过这个题? Yes 样例 给定一个二叉树: 1 / \ 2 3 / \ 4 5 返回其相同结构相同数值的克隆二叉 ...

  9. php 检查该数组有重复值

    if (count($array) != count(array_unique($array))) { echo '该数组有重复值'; }

  10. 停掉一台服务器,Nginx响应慢(转载)

    测试发现的问题及解决办法 1.当后端两台IIS应用服务器都正常时,访问速度非常快,查看日志,原来一个请求,是后端两台服务器同时响应的; 2.为了模仿故障测试,停掉一台IIS应用服务器,这时再访问,请求 ...