Game |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) |
| Total Submission(s): 50 Accepted Submission(s): 44 |
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Problem Description
Bob and Alice are playing a new game. There are n boxes which have been numbered from 1 to n. Each box is either empty or contains several cards. Bob and Alice move the cards in turn. In each turn the corresponding player should choose a non-empty box A and choose another box B that B<A && (A+B)%2=1 && (A+B)%3=0. Then, take an arbitrary number (but not zero) of cards from box A to box B. The last one who can do a legal move wins. Alice is the first player. Please predict who will win the game.
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Input
The first line contains an integer T (T<=100) indicating the number of test cases. The first line of each test case contains an integer n (1<=n<=10000). The second line has n integers which will not be bigger than 100. The i-th integer indicates the number of cards in the i-th box.
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Output
For each test case, print the case number and the winner's name in a single line. Follow the format of the sample output.
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Sample Input
2 |
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Sample Output
Case 1: Alice |
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Author
hanshuai@whu
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Source
The 5th Guangting Cup Central China Invitational Programming Contest
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Recommend
notonlysuccess
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/*
题意：1到n的盒子里，是空的或者装着卡片，每人轮流移动卡片，Alice先手，移动卡片的规则：每次选择一个A，然后选择一个B
要求 B<A&&(A+B)%2==1&&(A+B)%3==0,从A中抽取任意一张卡片放到B中，谁不能移动了，就输了。 初步思路：这个就是先将能移动的盒子对找出来，然后统计能移动的牌 #错误：上面的想法不完善，只是一方面的认为卡片会从一上一级向下一级移动。 #补充：阶梯博弈，实际上卡片是在3的余数上进行转移的。 #感悟：手残用了数组，开小了，wa了两发才发现
*/
#include<bits/stdc++.h>
using namespace std;
int n;
int x;
int res=;
int t;
void init(){
res=;
}
int main(){
// freopen("in.txt","r",stdin);
scanf("%d",&t);
for(int ca=;ca<=t;ca++){
init();
scanf("%d",&n);
for(int i=;i<=n;i++){
scanf("%d",&x);
if(i%==||i%==||i%==)
res^=x;
}
if(res) printf("Case %d: Alice\n",ca);
else printf("Case %d: Bob\n",ca);
}
return ;
}
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