Codeforces Round #416 (Div. 2) A. Vladik and Courtesy-优快云博客

本文介绍了一个简单的编程问题，即两个角色通过一系列递增数量的糖果交换来确定谁会首先无法完成下一轮交换。文章提供了完整的代码实现，并展示了如何通过输入不同数量的糖果来得到输出结果。

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Vladik and Courtesy
time limit per test 2 seconds
memory limit per test 256 megabytes
Description
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.

input
1 1
output
Valera
input
7 6
output
Vladik

水题啊，完全的水题啊ヾ(o◕∀◕)ﾉ
直接上代码

#include<bits/stdc++.h>
using namespace std;
int a,b;
int main()
{
    scanf("%d%d",&a,&b);
    long long int k=1;
    while(a>=0&&b>=0)
    {
        a-=k;
        k++;
        //printf("%d\n",k);
        b-=k;
        k++;
        //printf("%d\n",k);
    }
    if(a<0)
        printf("Vladik\n");
    else if(b<0)
        printf("Valera\n");
}