Codeforces Round #280 (Div. 2) D_html/css_WEB-ITnose

题目：

D. Vanya and Computer Game

time limit per test

memory limit per test

input

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.

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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 21234

output

VanyaVovaVanyaBoth

input

2 1 112

output

BothBoth

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.

分析：感觉题目描述不是很清楚。。sad。。。 读了好久。

第一反应，数据范围好大的样子。仔细分析发现在正整数秒的时候一定是both，而在整数秒间就是循环节了。

把第一秒的顺序打表出来，后面就用mod求解了。极限情况也是1e6，1s足够了。

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>#include <vector>#include <queue>#include <stack>#include <cassert>#include <algorithm>#include <cmath>#include <set>#include <limits>#include <map>using namespace std;#define MIN(a, b) ((a) < (b) ? (a) : (b))#define MAX(a, b) ((a) > (b) ? (a) : (b))#define F(i, n) for(int (i)=0;(i)<(n);++(i))#define REP(i, s, t) for(int (i)=(s);(i)<=(t);++(i))#define UREP(i, s, t) for(int (i)=(s);(i)>=(t);--(i))#define REPOK(i, s, t, o) for(int (i)=(s);(i)<=(t) && (o);++(i))#define MEM0(addr) memset((addr), 0, sizeof((addr)))#define PI 3.1415926535897932384626433832795#define HALF_PI 1.5707963267948966192313216916398#define MAXN 1000010;#define MAXM 10000#define MOD 1000000007typedef long long LL;const double maxdouble = numeric_limits<double>::max();const double eps = 1e-10;const int INF = 0x7FFFFFFF;string a[10000100];int main(){    int n;    double x,y;    int cnt1,cnt2,cnt;    while(cin >> n >> x>> y)    {        cnt = cnt1 = cnt2 = 1;        while(cnt1/x <= 1&&cnt2/y <= 1)        {            if(cnt1/x < cnt2/y)            {                a[cnt++] = "Vanya";                ++cnt1;            }else if(cnt1/x > cnt2/y){                a[cnt++] = "Vova";                ++cnt2;            }else            {                a[cnt++] = "Both";                a[cnt++] = "Both";                cnt1++,cnt2++;            }        }        while(n--)        {            int x;            scanf("%d",&x);            if(x%(cnt-1) == 0)                cout<<a[cnt-1]<<endl;            else                cout<<a[x%(cnt-1)]<<endl;        }    }    return 0;}