zoj3529

A Game Between Alice and Bob

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Time Limit: 5 Seconds Memory Limit: 262144 KB

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Alice and Bob play the following game. A series of numbers is written on the blackboard. Alice and Bob take turns choosing one of the numbers, and replace it with one of its positive factor but not itself. The one who makes the product of all numbers become 1 wins. You can assume Alice and Bob are intelligent enough and Alice take the first turn. The problem comes, who is the winner and which number is Alice's first choice if she wins?

Input

This problem contains multiple test cases. The first line of each case contains only one number N (1<= N <= 100000) representing there are N numbers on the blackboard. The second line contains N integer numbers specifying the N numbers written on the blackboard. All the numbers are positive and less than or equal to 5000000.

Output

Print exactly one line for each test case. The line begins with "Test #c: ", where c indicates the case number. Then print the name of the winner. If Alice wins, a number indicating her first choice is acquired, print its index after her name, separated by a space. If more than one number can be her first choice, make the index minimal.

Sample Input

4
5 7 9 12
4
41503 15991 72 16057 

Sample Output

Test #1: Alice 1
Test #2: Bob

题意：两个人博弈，给出n个数，每次可把一个数变成其因子，直到所有为1，无法再变者输。

思路：一个数可以分解成若干质因子乘积，n=p1^x1*p2^x2……pm^xm，则其至多能取的方案数就等于x1+x2+……xm，这就是n的sg值。

例如12=2*2*3，下面给出几个数据

sg[1]=0

sg[2]=sg[3]=mex{sg[1]}=1

sg[4]=mex{sg[1],sg[2]}=2

sg[6]=mex{sg[1],sg[2],sg[3]}=2

sg[12]=mex{sg[1],sg[2],sg[3],sg[6]}=3

可以看出sg[4]跟sg[6]是一样的，4想当于12取掉一个3变成2*2，6相当于12取掉2变成2*3，而4跟6最多还能取掉两个数才会变成1，因此可以得出质因子他们的个数和就是sg值

而题目后半部分，求第几个数，则根据剩余部分的总的异或值是否小于当前数的亦或值，若当前值大于剩余部分的异或值，相当于你可以把当前数，变成与剩余部分异或值相同，使得对手必败

#include<stdio.h>
#include<string.h>
int sg[5000005];
int NIM[100005];
bool isprime[5000005];
int prime[1000000],top;
void init(){
	memset(sg,-1,sizeof(sg));
	sg[1]=0;
	top=0;
	for(int i=2;i<=5000000;i++){
		if(!isprime[i]){
			prime[top++]=i;
			sg[i]=1;
			for(int j=i+i;j<=5000000;j+=i)
				isprime[j]=true;
		}
	}
}
int getsg(int num){
	if(sg[num]!=-1) return sg[num];
	int ans=0,temp=num;
	for(int i=0;prime[i]*prime[i]<=temp;i++)
		while(temp%prime[i]==0){
			ans++;
			temp/=prime[i];
		}
	if(temp!=1)
		ans++;
	return sg[num]=ans;
}
void slove(){
	int n;
	int ans,ti=1;
	while(~scanf("%d",&n)){
		ans=0;
		for(int i=1;i<=n;i++){
			scanf("%d",&NIM[i]);
			ans^=getsg(NIM[i]);
		}
		printf("Test #%d: ",ti++);
		if(ans==0)
			printf("Bob\n");
		else {
			printf("Alice ");
			for(int i=1;i<=n;i++)
				if((ans^sg[NIM[i]])<sg[NIM[i]]){
					printf("%d\n",i);
					break;
				}
		}
	}
}
int main(void){
	init();
	slove();
	return 0;
}