Codeforces Round #280 (Div. 2) D. Vanya and Computer Game 预处理
2 seconds
256 megabytes
standard input
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.
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.
#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[];
int main()
{
memset(dp,,sizeof(dp));
ll n,x,y;
cin>>n>>x>>y;
int flag2=;
int path1=;
int path2=;
int path=;
while()
{
if(path1==x&&path2==y)
break;
if(path1*y<path2*x)
{
dp[++path]=;
path1++;
}
else if(path1*y==path2*x)
{
dp[++path]=;
dp[++path]=;
path1++;
path2++;
}
else
{
dp[++path]=;
path2++;
}
//cout<<dp[path]<<endl;
}
while(n--)
{
ll shengming;
scanf("%I64d",&shengming);
shengming=shengming%(x+y);
if(shengming==x+y-||shengming==||dp[shengming]==)
printf("Both\n");
else if(dp[shengming]==)
{
if(flag2==)
printf("Vanya\n");
else
printf("Vova\n");
}
else
if(flag2==)
printf("Vova\n");
else
printf("Vanya\n");
}
return ;
}
Codeforces Round #280 (Div. 2) D. Vanya and Computer Game 预处理的更多相关文章
- 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 数学
D. Vanya and Computer Game time limit per test 2 seconds memory limit per test 256 megabytes input s ...
- Codeforces Round #280 (Div. 2) E. Vanya and Field 数学
E. Vanya and Field Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/492/pr ...
- Codeforces Round #280 (Div. 2) C. Vanya and Exams 贪心
C. Vanya and Exams Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/492/pr ...
- Codeforces Round #280 (Div. 2)E Vanya and Field(简单题)
转载请注明出处: http://www.cnblogs.com/fraud/ ——by fraud 本场题目都比较简单,故只写了E题. E. Vanya and Field Vany ...
- Codeforces Round #280 (Div. 2)_C. Vanya and Exams
C. Vanya and Exams time limit per test 1 second memory limit per test 256 megabytes input standard i ...
- Codeforces Round #280 (Div. 2) E. Vanya and Field 思维题
E. Vanya and Field time limit per test 2 seconds memory limit per test 256 megabytes input standard ...
- Codeforces Round #280 (Div. 2) A. Vanya and Cubes 水题
A. Vanya and Cubes time limit per test 1 second memory limit per test 256 megabytes input standard i ...
- 水题 Codeforces Round #308 (Div. 2) A. Vanya and Table
题目传送门 /* 水题:读懂题目就能做 */ #include <cstdio> #include <iostream> #include <algorithm> ...
随机推荐
- 我所知道的MVVM框架(转 司徒大大 )
RubyLouvre commented on 6 Sep 2014 avalon http://avalonjs.github.io/ (使用Object.defineProperties. V ...
- HTML5 之图片上传预处理
在开发 H5 应用的时候碰到一个问题,应用只需要一张小的缩略图,而用户用手机上传的确是一张大图,手机摄像机拍的图片好几 M,这可要浪费很多流量. 像我这么为用户着想的程序员,绝对不会让这种事情发生的, ...
- 洛谷P2822组合数问题
传送门啦 15分暴力,但看题解说暴力分有30分. 就是找到公式,然后套公式.. #include <iostream> #include <cstdio> #include & ...
- ZOJ 3962 Seven Segment Display(数位DP)
题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3962 题目大意: 有t组数据. 给你一个n,和8位的十六进制数s ...
- JS中精选this关键字的指向规律你记住了吗
1.首先要明确: 谁最终调用函数,this指向谁 this指向的永远只可能是对象!!!!! this指向谁永远不取决于this写在哪,而取 ...
- performance 判断页面是以哪种方式进入的
if (window.performance) { console.info("window.performance is supported"); console.log(per ...
- 用户说体验 | 关于阿里百川HotFix你需要了解的一些细节
最近很火的热修复技术,无意中了解到阿里百川也在做,而且Android.iOS两端都支持,所以决定试一试.试用一段时间后,感觉还不错,主要是他们有一个团队在不断维护更新这个产品,可以看到他们的版本更新记 ...
- Hive(四)Hive的3种连接方式与DbVisualizer连接Hive
一.CLI连接 进入到 bin 目录下,直接输入命令: [root@node21 ~]# hive SLF4J: Class path contains multiple SLF4J bindings ...
- 【LOJ】#2037. 「SHOI2015」脑洞治疗仪
题解 维护区间内1的个数,左边数0的长度,右边数0的长度,区间内0区间最长个数,覆盖标记 第一种操作区间覆盖0 第二种操作查询\([l_0,r_0]\)中1的个数,区间覆盖0,然后覆盖时找到相对应的区 ...
- 【51nod】1565 模糊搜索
题解 这个字符集很小,我们可以把每个字符拿出来做一次匹配,把第一个字符串处理每个出现过的该字符处理成一个区间加,即最后变成第一个字符串的该位置能够匹配某字符 例如对于样例 10 4 1 AGCAATT ...