Circle Game CodeForces - 1451D（博弈 +思维）-优快云博客

Utkarsh is forced to play yet another one of Ashish’s games. The game progresses turn by turn and as usual, Ashish moves first.

Consider the 2D plane. There is a token which is initially at (0,0)
. In one move a player must increase either the x coordinate or the y coordinate of the token by exactly k. In doing so, the player must ensure that the token stays within a (Euclidean) distance d from (0,0)
.

In other words, if after a move the coordinates of the token are (p,q)
, then p2+q2≤d2 must hold.

The game ends when a player is unable to make a move. It can be shown that the game will end in a finite number of moves. If both players play optimally, determine who will win.

Input

The first line contains a single integer t(1≤t≤100) — the number of test cases.

The only line of each test case contains two space separated integers d(1≤d≤105) and k (1≤k≤d ).

Output

For each test case, if Ashish wins the game, print “Ashish”, otherwise print “Utkarsh” (without the quotes).

Example

Input

5
2 1
5 2
10 3
25 4
15441 33

Output

Utkarsh
Ashish
Utkarsh
Utkarsh
Ashish

Note

In the first test case, one possible sequence of moves can be

(0,0)−→−−−Ashish (0,1)−→−−−−Utkarsh (0,2).

Ashish has no moves left, so Utkarsh wins.

在这里插入图片描述

题意：

每次走k步，求走d距离远（假如走的坐标是（x,y）,要时刻保证xx+yy<d*d）时谁能赢 Ashish先走

思路：

先计算出能走的最大s的值，保证ss+ss<d*d，s相当于是坐标x，然后再走，每走k步，步数++，最后判断k是否能整除2，如果能，则Ashish赢

代码：

#include<stdio.h>
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        long long d,k;
        scanf("%lld %lld",&d,&k);
        long long s=(d/k)*k;//求能整除k的数
        long long s1,s2=0;
        while(s*s+s*s>=d*d)
            s=s-k;      
        s1=s;
        while(s*s+s1*s1<=d*d)
        {
            s+=k;
            s2++;  //s2记录步数
        }
        if(s2%2==0)
            printf("Ashish\n");
        else
            printf("Utkarsh\n");
    }
    return 0;
}