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. hdu 3357 Stock Chase (图论froyd变形)

    Stock Chase Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total ...

  2. [bzoj3886] [USACO15JAN]电影移动Moovie Mooving

    题目链接 状压\(dp\). 注意到\(n\leq 20\)且每个只能用一次,所以很显然可以压缩每部电影看过没,记\(f[sta]\)为状态为\(sta\)时最多可以看多久. 转移时先枚举状态,然后枚 ...

  3. PyQt5学习--基本窗口控件--QMainWindow

    QMainWindow主窗口为用户提供一个应用程序框架,它有自己的布局,可以在布局中添加控件.比如将工具栏.菜单栏和状态栏等添加到布局管理器中. 窗口类型介绍 QMainWindow.QWidget和 ...

  4. [Leetcode] candy 糖果

    There are N children standing in a line. Each child is assigned a rating value. You are giving candi ...

  5. 遇到问题---java---git下载的maven项目web用tomcat发布时不带子项目

    遇到的情况是用git下载maven项目,然后用mvn eclipse:eclipse命令标记为eclipse项目之后,使用maven插件导入之后用tomcat发布运行,发现maven关联的几个子项目没 ...

  6. 1、linux下mysql5.5.20安装过程报错汇总

    1.Access denied for user 'root'@'localhost' (using password: YES) 这个提示是因为root帐户默认不开放远程访问权限,所以需要修改一下相 ...

  7. 群联MPALL(Rel) 7F V5.03.0A-DL07量产工具 PS2251-07(PS2307)

    前言:U盘被写保护,真的很醉人啊~~      群联MPALL是一款群联PS2251系列主控量产修复工具,本版本支持PS2251-67.PS2251-68.PS2251-02.PS2251-03.PS ...

  8. 微信小程序基础知识

    一.基本目录结构 app.js 定义app入口 app.json 定义页面配置 index.js 页面中的事件和监听 index.wxml 定义布局文件 1.app.json配置基本信息 { “pag ...

  9. C# 序列化理解 2(转)

    一.概述 序列化是把对象转变成流.相反的过程就是反序列化. 哪些场合用到这项技术呢? 1. 把对象保存到本地,下次运行程序时恢复这个对象. 2. 把对象传送到网络的另一台终端上,然后在此终端还原这个对 ...

  10. ACM-ICPC 2018 南京赛区网络预赛 Sum

    A square-free integer is an integer which is indivisible by any square number except 11. For example ...