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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.
Single line of input data contains two space-separated integers a, b (1 ≤ a, b ≤ 109) — number of Vladik and Valera candies respectively.
Pring a single line "Vladik’’ in case, if Vladik first who can’t give right amount of candy, or "Valera’’ otherwise.
1 1
Valera
7 6
Vladik
Illustration for first test case:
Illustration for second test case:
题意：
很水的题目，注意的是别人给的不能算自己的。
代码：
#include#include #include using namespace std;int main(){ int a,b,i; while(cin>>a>>b) { for(i=1;;i++) { if(i%2==0) { b=b-i; if(b<0){cout<<"Valera"<
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