Time Limit:5000MS Memory Limit:262144KB 64bit IO Format:%lld
& %llu
Description
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
题目大意：
给定n个数字。两个人轮流玩游戏。把这n个数字变成n的因子（不包括本身），最后全变成1就赢了，问你谁会赢，假设Alice赢了，把第一步选择的那一个数字的序号输出，假设有多种方案，输出序号小的。
解题思路：
sg[1]=0;
sg[2]=mex{sg[1]}=1;
sg[3]=mex{sg[1]}=1;
sg[4]=mex{sg[2],sg[1]}=2;
sg[5]=mex{sg[1]}=1;
sg[6]=mex{sg[2],sg[3]}=2;
sg[7]=mex{sg[1],sg[7]}=1;
sg[8]=mex{sg[1],sg[2],sg[4]}=3;
..........................
发现 sg[x]为 x因子的个数。证明：由于 假设一个为a, 则：sg[x]=sg[(x/a)*a]=mex{sg[1]....,sg[(x/a)]}=sg[(x/a)]+1;
求出每一个数字的sg后，仅仅须要看sg的异或和，假设==0 和明显，输出Bob
否则。输出Alice，可是要输出Alice先走了哪一步，这里实用到了一个性质：仅仅须要用合并后的SG值与每一堆SG值分别异或，看得到的结果是否小于原来该堆的SG值。假设小则能够取该堆。
这个性质临时不是非常懂，先用上，以后了解了会给出证明和过程。
解题代码：
#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
using namespace std; const int maxn=5000010;
int n,sg[maxn];
int d[110000]; vector <int> prime;
bool isprime[maxn]; void ini(){
isprime[2]=true;
for(int i=3;i<maxn;i+=2) isprime[i]=true;
for(int i=3;i<maxn;i+=2){
for(int j=i;j<maxn/i;j+=2){
isprime[i*j]=false;
}
}
for(int i=2;i<=maxn;i++){
if(isprime[i]) prime.push_back(i);
}
} int SG(int x){
if(sg[x]!=-1) return sg[x];
int ret=0,size=prime.size(),tmp=x;
for(int i=0;i<size && prime[i]*prime[i]<=x;i++){
while(x%prime[i]==0){
x/=prime[i];
ret++;
}
}
if(x>1) ret++;
return sg[tmp]=ret;
} int main(){
ini();
memset(sg,-1,sizeof(sg));
int casen=0;
while(scanf("%d",&n)!=EOF){
int ans=0;
for(int i=0;i<n;i++){
scanf("%d",&d[i]);
ans^=SG(d[i]);
}
if(ans==0) printf("Test #%d: Bob\n",++casen);
else{
for(int i=0;i<n;i++){
if( ( ans^SG(d[i]) ) < SG(d[i]) ){
printf("Test #%d: Alice %d\n",++casen,i+1);
break;
}
}
}
}
return 0;
}