Little Q and Little T are playing a game on a tree. There are
n![技术分享图片]()
vertices on the tree, labeled by 1,2,...,n![技术分享图片]()
, connected by n?1![技术分享图片]()
bidirectional edges. The i![技术分享图片]()
-th vertex has the value of w![技术分享图片]()
i![技术分享图片]()
![技术分享图片]()
.
In this game, Little Q needs to grab some vertices on the tree. He can select any number of vertices to grab, but he is not allowed to grab both vertices that are adjacent on the tree. That is, if there is an edge between x![技术分享图片]()
and y![技术分享图片]()
, he can‘t grab both x![技术分享图片]()
and y![技术分享图片]()
. After Q‘s move, Little T will grab all of the rest vertices. So when the game finishes, every vertex will be occupied by either Q or T.
The final score of each player is the bitwise XOR sum of his choosen vertices‘ value. The one who has the higher score will win the game. It is also possible for the game to end in a draw. Assume they all will play optimally, please write a program to predict the result.
The first line of the input contains an integer
T(1≤T≤20)![技术分享图片]()
, denoting the number of test cases.
In each test case, there is one integer n(1≤n≤100000)![技术分享图片]()
in the first line, denoting the number of vertices.
In the next line, there are n![技术分享图片]()
integers w![技术分享图片]()
1![技术分享图片]()
,w![技术分享图片]()
2![技术分享图片]()
,...,w![技术分享图片]()
n![技术分享图片]()
(1≤w![技术分享图片]()
i![技术分享图片]()
≤10![技术分享图片]()
9![技术分享图片]()
)![技术分享图片]()
, denoting the value of each vertex.
For the next n?1![技术分享图片]()
lines, each line contains two integers u![技术分享图片]()
and v![技术分享图片]()
, denoting a bidirectional edge between vertex u![技术分享图片]()
and v![技术分享图片]()
.
For each test case, print a single line containing a word, denoting the result. If Q wins, please print Q. If T wins, please print T. And if the game ends in a draw, please print D.
Q
题解：由于是求异或，我们只需考虑每一位上的 1 的个数即可，比如：对于最高位，如果为偶数，那么小Q只需选一个或者不选即可那么小Q，小T最后该位上的数是相同的，如果为奇数，小Q选一个为必胜，因为小Q就选这一个，剩下都给小T，小T的最终结果必定小于小Q， 因此，我们只需要对每一位上的1的个数加一遍，如果有出现奇数个，则Q必胜，否者平局；
参考代码为：
#include<bits/stdc++.h>
using namespace std;
const int maxn=1e5+5;
int w[maxn],u[maxn],v[maxn];
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
int T,n;
cin>>T;
while(T--)
{
cin>>n;
bool flag=false;
for(int i=1;i<=n;i++) cin>>w[i];
for(int i=1;i<n;i++) cin>>u[i]>>v[i];
for(int i=0;i<32;i++)
{
int cnt=0;
for(int i=1;i<=n;i++)
{
if(w[i]&1) cnt++;
w[i]>>=1;
}
if(cnt & 1)
{
cout<<"Q"<<endl;
flag=true;
break;
}
}
if(!flag) cout<<"D"<<endl;
}
return 0;
}
