codeforces #416 div2 811A-Vladik and Courtesy-优快云博客

本文介绍了一种通过算法判断糖果交换游戏中谁会首先无法继续游戏的方法。游戏中两名玩家轮流互相给予糖果，每次给予的数量比上一次多一个。文章提供了一个C++实现方案，通过计算每个玩家能够继续游戏的最大糖果数量来确定胜者。

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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:

$\text{[math]}$

一道水题，但可能想复杂了。

#include<cstdio>
#include<algorithm>
#define ll long long 
using namespace std;
int main() {
	int a, b;
	scanf("%d %d", &a, &b);
	ll s1,s2,t1,t2;
	int h, k;
	int flag1 = 0;
	int flag2 = 0;
	for (int i = 0;i < 1e5;i++) {
		s1 = i*i;
		s2 = i*(i + 1);
		if (s1 >= a && !flag1) {
			flag1 = 1;
			h = i;
			t1 = s1;
		}
		if (s2 >= b && !flag2) {
			flag2 = 1;
			k = i;
			t2 = s2;
		}
		if (flag1&&flag2) break;
	}
	if (t1 == a&&t2 == b) {
		if (h > k) printf("Valera\n");
		else printf("Vladik\n");
	}
	if (t1 > a&&t2 > b) {
		if (h > k) printf("Valera\n");
		else printf("Vladik\n");
	}
	if (t1 == a&&t2 > b) {
		if (k >= h + 1) printf("Vladik\n");
		else printf("Valera\n");
	}
	if (t1 > a&&t2 == b) {
		if (h - 1 <= k) printf("Vladik\n");
		else printf("Valera\n");
	}
}