cf492D Vanya and Computer Game
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.
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.
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.
- 4 3 2
1
2
3
4
- Vanya
Vova
Vanya
Both
- 2 1 1
1
2
- Both
Both
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.
前三题都是无脑题……
但是第四题也是吧……
题意是第一个人每秒开x枪,第二个人每秒开y枪,下面给出n个询问,问boss血量ai的时候谁打出最后一下
首先x、y约分。因为题目只要求最后的结果,那么假设x=g*x0,y=g*y0,g>1
那么问题可以转换为“第一个人每1/g秒开x0枪,第二个人每1/g秒开y0枪”
实际上是一样的
那么直接打表暴力搞出1~x0+y0的胜负情况,然后O(1)输出
因为数组开小而RE3次……蛋疼
- #include<cstdio>
- #include<iostream>
- #include<cstring>
- #include<cstdlib>
- #include<algorithm>
- #include<cmath>
- #include<queue>
- #include<deque>
- #include<set>
- #include<map>
- #include<ctime>
- #define LL long long
- #define inf 0x7ffffff
- #define pa pair<int,int>
- #define pi 3.1415926535897932384626433832795028841971
- using namespace std;
- inline LL read()
- {
- LL x=0,f=1;char ch=getchar();
- while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
- while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
- return x*f;
- }
- int n,nx=1,ny=1;
- int mod,x,y;
- int ans[3000010];
- inline int gcd(int a,int b)
- {return b==0?a:gcd(b,a%b);}
- int main()
- {
- n=read();x=read();y=read();
- int g=gcd(x,y);
- x/=g;y/=g;
- mod=x+y;
- for (int i=1;i<=mod;i++)
- {
- LL aa=(LL)nx*y,bb=(LL)ny*x;
- if (aa<bb && nx<=x)ans[i]=1,nx++;
- else if (aa>bb && ny<=y)ans[i]=-1,ny++;
- else
- {
- i++;
- nx++;
- ny++;
- }
- }
- ans[0]=ans[mod];
- for (int i=1;i<=n;i++)
- {
- int query=read();
- query%=mod;
- if (ans[query]==1)printf("Vanya\n");
- else if (ans[query]==-1)printf("Vova\n");
- else printf("Both\n");
- }
- return 0;
- }
cf492D Vanya and Computer Game的更多相关文章
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 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 ...
- 【cf492】D. Vanya and Computer Game(二分)
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 ...
- 【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 ...
- 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
- java中 SSL认证和keystore使用
java中 SSL认证和keystore使用 2013-10-12 11:08 10488人阅读 评论(0) 收藏 举报 目录(?)[+] 好久没用过SSL认证了,东西久不用,就有点生疏. ...
- 自定义input file样式
自定义input file样式:一般都是通过隐藏input,通过定义label来实现.这种做法要注意的是label的for属性要指定input对应的id; <!DOCTYPE html> ...
- C++编程规范之12:懂得何时和如何进行并发性编程
摘要: 如果应用程序使用了多个线程或者进程,应该知道如何尽量减少共享对象,以及如何安全地共享必须共享的对象. 在多线程和并发编程中最重要的是要避免死锁.活锁和恶性的竞争条件. 在编写多线程程序时要注意 ...
- swift 创建tableView 并实现协议
import UIKit class ViewController2: UIViewController,UITableViewDelegate,UITableViewDataSource{ ...
- Intel 被 ARM 逼急了
英特尔最近推出基于Silvermont架构Bay Trail系列处理器,相对前一代Bonnell架构的最突出的改进就是支持乱序执行 silvermon架构的处理器将出现在pc,平板等: List of ...
- UIView层次管理bringSubviewToFront,sendSubviewToBack
将一个UIView显示在最前面只需要调用其父视图的 bringSubviewToFront()方法. 将一个UIView层推送到背后只需要调用其父视图的 sendSubviewToBack()方法. ...
- Movie播放Gif,完美实现屏幕适配
android播放gif 我研究过3种 第一 :GifView支持android播放gif,效果是 先加载第一帧,然后慢慢加载完其他的针,这样效果视觉很不好,是从模糊到清晰的过程:第二:是流行的把g ...
- canvas--画宇宙
<!doctype html><html lang="en"> <head> <meta charset="UTF-8" ...
- 为什么class中属性以空格分隔?
1 div.contain .blue{color:blue;}/*后代选择器*/2 div.contain.blue{color:blue;} /*多类选择器*/ 以上两种规则分别应用的元素如下: ...
- a标签的onclick和href事件的区别
在执行顺序上href是低于onclick的,那么这个会造成什么影响呢 <div onclick="a()"> <a href="#" oncl ...