zoukankan      html  css  js  c++  java
  • 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
    ");
            else if(dp[shengming]==1)
            {
                if(flag2==0)
                    printf("Vanya
    ");
                else
                    printf("Vova
    ");
            }
            else
                if(flag2==0)
                    printf("Vova
    ");
                else
                    printf("Vanya
    ");
        }
        return 0;
    }
  • 相关阅读:
    知识搜索
    使用 getopt() 进行命令行处理
    【新提醒】夏新大v安卓4.1尝鲜最新更新版本发布(包含进步版)1124更新 大V综合交流区 360论坛
    搜狗知立方高调亮相 开启知识计算新时代
    socat: Linux / UNIX TCP Port Forwarder
    Crontab 每两周执行一次
    python 命令行解析 optionparser
    crontab jojo's blog--快乐忧伤都与你同在 BlogJava
    搜索引擎开始「实体搜索」新领域竞争,Google、百度分别发力实体搜索产品
    netcat(nc)的替代产品 Socat
  • 原文地址:https://www.cnblogs.com/qscqesze/p/4136599.html
Copyright © 2011-2022 走看看