空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld
题目描述
There is a game consisted of two players Alice and Bob, and one referee Jelly.
At first, Jelly will choose a special integer X and write down on the paper, but don’t show it to each player. Then Jelly will give players two integers L and R, which means X ranges from L to R inclusively. Alice and Bob start to guess the integer X in turns, and Alice goes first. The first one who guesses the integer X loses the game.
The specific rules of the game are as follows:
One player guesses an integer K (K ∈ [L, R]), and Jelly will announce one of three results:
1. “Lose”. That means K is equal to X, this player loses the game.
2. “Large”. That means K is larger than X , other one should guess the integer from L to K − 1 inclusively in the next. (R becomes K − 1)
3. “Small.” That means K is smaller than X, other one should guess the integer from K+1 to R inclusively in the next. (L becomes K + 1)
If one player loses the game, Jelly will show players the paper with X so as to prevent Jelly to cheat the players.
One day, Alice and Bob play this game again. But they reach a consensus that the loser needs to pay for the winner’s shopping cart. Cause neither of them wants to lose the game, they order Braised Chicken and Rice(HuangMenJi) for Jelly at different times. In return, Jelly tells Alice and Bob the special integer X secretly. So, both players will use the best strategy to play the game.
Now, the question is who will win the game under this situation. (Alice goes first)
输入描述:
• 1≤L≤X≤R≤105.
输出描述:
For each test case, output one line containing “Case #x: y”, where x is the test case number (starting
from 1) and y is the winner’s name.
输入
2
1 1 1
1 2 2
输出
Case #1: Bob
Case #2: Alice
备注：
For the first case, Alice can only guess the integer 1, so she loses the game, the winner is Bob.
For the second case, if Alice guesses the integer 1, then Bob can only guess the integer 2. So the winner
is Alice.
给定一个闭区间L~R和一个在区间内的数字X，Bob和Alice猜数字，如果所猜的数字K=X，则输；如果K>X,则区间变为L~K-1；如果K<X，则区间变为K+1~R，输出输者
Alice先猜并且两人都知道X的数值
取胜策略：使X两端拥有相同多的数字
当为a x b形式时下一个猜数字的人一定输
#include<iostream>
using namespace std;
int main()
{
int t,l,r,x;
cin>>t;
for(int i=0;i<t;i++)
{
cin>>l>>r>>x;
if((x-l)==(r-x))cout<<"Case #"<<i+1<<": Bob"<<endl;//Bob先使x两端长度相同
else cout<<"Case #"<<i+1<<": Alice"<<endl;//Alice先使x两端数字相同
}
return 0;
}