time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

At regular competition Vladik and Valera won a and b candies respectively. Vladik offered 1 his candy to Valera. After that Valera gave Vladik 2 his candies, so that no one thought that he was less generous. Vladik for same reason gave 3 candies to Valera in next turn.

More formally, the guys take turns giving each other one candy more than they received in the previous turn.

This continued until the moment when one of them couldn’t give the right amount of candy. Candies, which guys got from each other, they don’t consider as their own. You need to know, who is the first who can’t give the right amount of candy.

Input

Single line of input data contains two space-separated integers a, b (1 ≤ a, b ≤ 109) — number of Vladik and Valera candies respectively.

Output

Pring a single line "Vladik’’ in case, if Vladik first who can’t give right amount of candy, or "Valera’’ otherwise.

Examples
Input
1 1
Output
Valera
Input
7 6
Output
Vladik
Note

Illustration for first test case:

Illustration for second test case:

题意就是两个人依次拿出多于上一个人的糖。

代码1:

#include<bits/stdc++.h>
using namespace std;
int main(){
int n,m;
int num1,num2;
while(~scanf("%d%d",&n,&m)){
num1=;num2=;
int t=n;
for(int i=;i<=t;i=i+){
if(n>=i+){
  num1++;
  n=n-(i+);
}
if(m>=i+){
  num2++;
   m=m-(+i);
}
}
//cout<<num1<<" "<<num2<<endl;
if(num1<=num2) printf("Vladik\n");
else printf("Valera\n");
}
return ;
}

代码2:

#include<bits/stdc++.h>
using namespace std;
int main(){
int n,m;
int num1,num2;
while(~scanf("%d%d",&n,&m)){
num1=;num2=;
int t=n;
for(int i=;i<=t;i++){ //我一开始写的i=0;i<=n,n会变的。
if(n>=*i-){
num1++;
n-=*i-;
}
if(m>=*i){
num2++;
m-=*i;
}
}
if(num1<=num2) printf("Vladik\n");
else printf("Valera\n");
}
return ;
}

Codeforces 811 A. Vladik and Courtesy的更多相关文章

  1. codeforces 811 D. Vladik and Favorite Game(bfs水题)

    题目链接:http://codeforces.com/contest/811/problem/D 题意:现在给你一个n*m大小的图,你输出一个方向之后,系统反馈给你一个坐标,表示走完这步之后到的位子, ...

  2. codeforces 811 E. Vladik and Entertaining Flags(线段树+并查集)

    题目链接:http://codeforces.com/contest/811/problem/E 题意:给定一个行数为10 列数10w的矩阵,每个方块是一个整数, 给定l和r 求范围内的联通块数量 所 ...

  3. codeforces 811 C. Vladik and Memorable Trip(dp)

    题目链接:http://codeforces.com/contest/811/problem/C 题意:给你n个数,现在让你选一些区间出来,对于每个区间中的每一种数,全部都要出现在这个区间. 每个区间 ...

  4. Codeforces 811 C. Vladik and Memorable Trip

    C. Vladik and Memorable Trip   time limit per test 2 seconds memory limit per test 256 megabytes inp ...

  5. Codeforces 811 B. Vladik and Complicated Book

    B. Vladik and Complicated Book   time limit per test 2 seconds memory limit per test 256 megabytes i ...

  6. Codeforces Round #416 (Div. 2) A. Vladik and Courtesy【思维/模拟】

    A. Vladik and Courtesy time limit per test 2 seconds memory limit per test 256 megabytes input stand ...

  7. 【35.02%】【codeforces 734A】Vladik and flights

    time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard o ...

  8. 【codeforces 743E】Vladik and cards

    [题目链接]:http://codeforces.com/problemset/problem/743/E [题意] 给你n个数字; 这些数字都是1到8范围内的整数; 然后让你从中选出一个最长的子列; ...

  9. 【44.64%】【codeforces 743C】Vladik and fractions

    time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard ou ...

随机推荐

  1. 为Ubuntu安装FTP服务

    打开"终端窗口",输入"sudo apt-get update"-->回车-->"输入当前登录用户的管理员密码"-->回车 ...

  2. request.getParameterMap() 获取表单提交的键值对 并且 也能获取动态表单的key

    Map<String,String[]> map = request.getParameterMap();Set<String> keys = map.keySet(); 获取 ...

  3. P1886 滑动窗口

    题目描述 现在有一堆数字共N个数字(N<=10^6),以及一个大小为k的窗口.现在这个从左边开始向右滑动,每次滑动一个单位,求出每次滑动后窗口中的最大值和最小值. 例如: The array i ...

  4. [BZOJ3196][Tyvj1730]二逼平衡树

    [BZOJ3196][Tyvj1730]二逼平衡树 试题描述 您需要写一种数据结构(可参考题目标题),来维护一个有序数列,其中需要提供以下操作: 查询 \(k\) 在区间内的排名 查询区间内排名为 \ ...

  5. wait for it

  6. BZOJ day2

    十六题...(好难啊) 1051105910881191119214321876195119682242243824562463276128184720

  7. CVPR2014 Objectness 源码转换(完整版) VS2012 X64 –>win32

    一.版本转换  1.将源码中vs2012 X64版本转换为vs2012 win32版本. 2.源码下载及其相关资料下载http://mmcheng.net/zh/bing/ 3.需要下载源码(Pape ...

  8. URAL1277 Cops and Thieves(最小割)

    Cops and Thieves Description: The Galaxy Police (Galaxpol) found out that a notorious gang of thieve ...

  9. 用实例工厂的方法实例化bean

    在实例化bean时,除了setter,constructor方法外,还有实例工厂方法,和静态工厂方法. 看代码: People类的代码如下: package com.timo.domain; publ ...

  10. eclipse集成mybatis的generater插件

    mybatis也能方向生成代码,能方向生成实体类(po).mapper接口和Mapper接口映射文件,能减少我们代码的工作量.详细步骤如下 1.下载mybatis生成架包工具MyBatis_Gener ...