博弈论2

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:

The first line contains an integer T, where T is the number of test cases. T test cases follow. 

For each test case, the only line contains three integers L, R and X.

• 1 ≤ T ≤ 10 ⁵.

• 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.

示例1

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.

  由X向两边扩散。 谁距离最先距离X近谁输。所以就判断一开始L,R距离X的距离如果距离相等则Bob胜，否则Alice胜

#include <iostream>
#include <algorithm>
#include <string.h>
#include <stdio.h>
using namespace std;
 
int main() 
{ 
    int ans,n,m,cnt,num; 
    scanf("%d",&n); 
    int cas=1;
    while(n--) 
    { 
         
        int l,r,p;
        cin>>l>>r>>p;
        if(r-p==p-l)
           printf("Case #%d: Bob\n",cas++); 
        else 
           printf("Case #%d: Alice\n",cas++); 
    }  
    return 0; 
}