Codeforces Round #416 (Div. 2) A. Vladik and Courtesy【思维/模拟】-优快云博客

本文介绍了一个有趣的编程问题，即两个朋友通过一系列递增数量的糖果交换来比拼慷慨度，直至一方无法继续。文章提供了问题背景、输入输出示例及解析，并附带了解决该问题的C++代码。

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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:

$\text{[math]}$

Illustration for second test case:

$\text{[math]}$

【题意】：看hint就懂了。

【分析】：根据天数的奇偶区分谁做东（减改天数）

【代码】：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define N 65535+10
int main()
{
    ll a,b;
    while(cin>>a>>b)
    {
       for(int i=1;i<=10000007;i++)
       {
           if(i%2==1)
           {
               a -= i;
               if(a<0) return 0*puts("Vladik");
           }
           else
           {
               b -= i;
               if(b<0) return 0*puts("Valera");
           }
       }
    }
}

转载于:https://www.cnblogs.com/Roni-i/p/8011968.html