Alice's Game - HDU 3544 博弈

Alice's Game

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 511    Accepted Submission(s): 211

Problem Description

Alice and Bob have got a lot of chocolates. All the chocolates are rectangles of different shapes as X_i * Y_i.They decide to play an interesting game on the chocolates. They take turns choose a chocolate and split it into two pieces. The one who can not take operations lose. Due to the game is too simple, the additional rules apply. Alice is only allowed to split the chocolate vertically, and Bob is only allowed to split the chocolate horizontally.
Specifically, for Alice, a chocolate X_i * Y_i, can only split into A * Y_i, and B * Y_i where A + B = X_i and A, B > 0. And for Bob, a chocolate X_i * Y_i, can only split into X_i * A, and X_i * B where A + B = Y_i and A, B > 0.
Alice and Bob are clever enough to take the optimal operation, if Alice plays first, your are to decide who will win.

 

Input

The input contains multiple test cases. The first line of input contains a single integer denoting the number of test cases.
For each test case, the first line contains an integer N, the number of pieces of chocolates. (1 <= N <= 100)
For the next N lines, each line contains two integers X_i and Y_i, denoting the chocolate sized X_i * Y_i. (1 <= X_i, Y_i <= 1000000000)

 

Output

For each test cases, output "Alice" when Alice will win, and "Bob" otherwise. See sample test cases for further details.

 

Sample Input

411 112 122 22 113 2

 

Sample Output

Case 1: BobCase 2: AliceCase 3: AliceCase 4: Bob

 

思路：同样不理解题目，本来想的是每次从中间分，但是怕超时，后来去网上看别人的思路，发现写的都是每次取另一个人切完的小的那块继续分。问题是题目中只说把一块切成两块，并没有说要丢掉一块啊。。。。那这样的话题意只能理解成第二个人可以选择新切成的两块中的一块并吃掉另一块。。。

AC代码如下：

#include<cstdio>#include<cstring>using namespace std;typedef long long ll;int T,t,n,m;int main(){    int i,j,k;    ll x,y,a,b;    scanf("%d",&T);    for(t=1;t<=T;t++)    {        scanf("%d",&n);        a=b=0;        for(i=1;i<=n;i++)        {            scanf("%I64d%I64d",&x,&y);            while(x>1 && y>1)            {                x/=2;y/=2;            }            if(y==1)              a+=x-1;            if(x==1)              b+=y-1;        }        printf("Case %d: ",t);        if(a>b)          printf("Alice\n");        else          printf("Bob\n");    }}