题目
Now it's time for lunch. Today's menu is chocolate!
Though every baby likes chocolate, the appetites of babies are little. After lunch, there are still n pieces of chocolate remained: The length of the ith piece is li.
Using the remained chocolate, Baby Volcano is going to play a game with his teacher, Mr. Sprague. The rule of the game is quite simple.
Two player plays in turns, and Baby Volcano will play first:
- In each turn, the player needs to select one piece of chocolate. If the length of the selected piece is equal to 1, the player of this turn will lose immediately.
- Suppose the length of the selected piece is l. Then the player needs to select a positive integer k satisfying k is at least 2 and k is a factor of l.
- Then the player needs to cut the selected piece into k pieces with length lk.
The game continues until one player selects a piece of chocolate with length 1.
Suppose both players plays optimally, your task is to determine whether Baby Volcano will win.
Input
The first line contains single integer t(1≤t≤2∗104), the number of testcases.
For each testcase, the first line contains a single integer n(1≤n≤10).
The second line contains n positive integers li(1≤li≤109), representing the length of each piece.
Output
For each testcase, output char 'W' if Baby Volcano will win, otherwise output char 'L'.
Sample Input
3
2
4 9
2
2 3
3
3 9 27
Sample Output
W
L
L
思路
首先，我们发现题目时间给了三秒，然后我们发现所有数都是质数的情况下，比赛的胜负就十分特殊了，它由数的奇偶性决定，所以我们猜想这题的做法大概率是要去做素数筛法的。所以特殊的当这个数为2的次方的时候，就形成了一种平衡局面，先手必胜，可操作次数就相当于是奇数质因子之和加上是否为偶数，然后做nim和就可以了。
代码实现
#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
#define rep(i,f_start,f_end) for (int i=f_start;i<=f_end;++i)
#define per(i,n,a) for (int i=n;i>=a;i--)
#define MT(x,i) memset(x,i,sizeof(x) )
#define rev(i,start,end) for (int i=start;i<end;i++)
#define inf 0x3f3f3f3f
#define mp(x,y) make_pair(x,y)
#define lowbit(x) (x&-x)
#define MOD 1000000007
#define exp 1e-8
#define N 1000005
#define fi first
#define se second
#define pb push_back
typedef long long ll;
const ll INF=0x3f3f3f3f3f3f3f3f;
typedef vector <int> VI;
typedef pair<int ,int> PII;
typedef pair<int ,PII> PIII;
ll gcd(ll a,ll b){return b==0?a:gcd(b,a%b);}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;};
void check_max (int &a,int b) { a=max (a,b);}
void check_min (int &a,int b) { a=min (a,b);}
inline int read() {
char ch=getchar(); int x=0, f=1;
while(ch<'0'||ch>'9') {
if(ch=='-') f=-1;
ch=getchar();
} while('0'<=ch&&ch<='9') {
x=x*10+ch-'0';
ch=getchar();
} return x*f;
}
const int maxn=sqrt (1e9)+5;
int mark[maxn],pri[maxn];
int t,ans,cnt=0;
int F (int x) {
if (x==1) return 0;
int c=x%2==0;
while (x%2==0) x/=2;
int g=sqrt (x);
for (int i=1;pri[i]<=g;i++) {
if (x%pri[i]==0) {
c++;
while ((x/=pri[i])%pri[i]==0) c++;
g=sqrt (x);
}
}
return (x>1)+c;
}
int main () {
rev (i,2,maxn) if (!mark[i]) {
pri[++cnt]=i;
for (int j=i+i;j<maxn;j+=i) mark[j]=1;
}
pri[++cnt]=1e6;
int t;
scanf ("%d",&t);
while (t--) {
int n,ans=0;
scanf ("%d",&n);
rep (i,1,n) {
int x;
scanf ("%d",&x);
ans^=F (x);
}
if (ans==0) printf ("L
");
else printf ("W
");
}
return 0;
}