Alice and Bob |
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| Accepted : 133 | Submit : 268 | |
| Time Limit : 1000 MS | Memory Limit : 65536 KB | |
Problem DescriptionThe famous "Alice and Bob" are playing a game again. So now comes the new problem which need a person smart as you to decide the winner. The problem is as follows: They are playing on a rectangle paper, Alice and Bob take turn alternatively, for each turn, InputFirst Line contains an integer t indicate there are t cases(1≤t≤1000) For each case: The input consists of two integers w and h(1≤w,h≤1,000,000,000), the size of rectangle. OutputFirst output Case number For each case output Alice or Bob, indicate the winner. Sample Input2 1 2 2 2 Sample OutputCase 1: Alice Case 2: Bob |
题意：一块长方形纸张。Alice先对折剪下，再由Bob对折剪下。若一个人无论怎么剪面积都是小数的时候。这个人就输了这场比赛。
思路：长方形的面积是长乘以宽，无论以哪一个边对折剪下，那条边都是除以2，若有一条边长度变为小数，则这个人就输了。假设模拟。有可能会超时，运用边找规律。
当两条边都变成奇数的时候，则这个人就输了。也就是说。看偶数能对折几次。
Alice and Bob
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
using namespace std;
int main()
{
int n,m,i,j,k,a,b;
scanf("%d",&n);
for(m = 1;m <= n;m++)
{
scanf("%d%d",&a,&b);
i = 0;
while(a % 2 == 0)
{
i++;
a = a / 2;
}
while(b % 2 == 0)
{
i++;
b = b / 2;
}
if(i % 2 == 0)
printf("Case %d: Bob\n",m);
else
printf("Case %d: Alice\n",m);
}
return 0;
}