Codeforces Round #416 (Div. 2) A. Vladik and Courtesy

本文介绍了一个有趣的糖果交换竞赛问题,两人轮流互相给予糖果,每人送出的糖果数依次递增,直至一方无法继续送出为止。文章提供了详细的算法分析及C语言实现。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

A. Vladik and Courtesy
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:有两个人,每次失去的比对方多1,问谁第一个无法满足。


分析:分别计算每个人能抗几轮,然后比较就可以了。


#include<stdio.h>
#include<string.h>
int main()
{
	long long a,b;
	scanf("%lld%lld",&a,&b);
	int ta=0,tb=0;
	int i=1;
	while(a-i>=0)
	{
		a-=i;
		ta++;
		i+=2;
	}
	i=2;
	while(b-i>=0)
	{
		b-=i;
		tb++;
		i+=2;
	}
	if(ta>tb)
		printf("Valera\n");
	else
	printf("Vladik\n");
	
}

评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值