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  • 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;
    }
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  • 原文地址:https://www.cnblogs.com/qscqesze/p/4136599.html
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