标签：mit java ini des nta ble href oev sea
Paint Chain
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2465 Accepted Submission(s): 880
Problem Description
Aekdycoin and abcdxyzk are playing a game. They get a circle chain with some beads. Initially none of the beads is painted. They take turns to paint the chain. In Each turn one player must paint a unpainted beads. Whoever is unable to paint in his turn lose the game. Aekdycoin will take the first move.
Now, they thought this game is too simple, and they want to change some rules. In each turn one player must select a certain number of consecutive unpainted beads to paint. The other rules is The same as the original. Who will win under the rules ?You may assume that both of them are so clever.
Input
First line contains T, the number of test cases. Following T line contain 2 integer N, M, indicate the chain has N beads, and each turn one player must paint M consecutive beads. (1 <= N, M <= 1000)
Output
For each case, print "Case #idx: " first where idx is the case number start from 1, and the name of the winner.
Sample Input
Sample Output
Case #1: aekdycoin
Case #2: abcdxyzk
Author
jayi
Source
一串长度为n的圆形珠子，每次可以找连续的m个白色珠子染色，不能操作者输，问先手状态。
先计算出线性状态下的游戏的sg函数，再计算与真实状态的sg函数时判断sg[n-m]=sg[n-m]==0?1:0。
由于子状态只有一种。
1 #include<bits/stdc++.h>
2 using namespace std;
3 int sg[1010];
4 bool vis[1010];
5 int main(){
6 int t,n,m,i,j,k;
7 cin>>t;
8 for(int cas=1;cas<=t;++cas){
9 cin>>n>>m;
10 printf("Case #%d: ",cas);
11 if(n<m){
12 puts("abcdxyzk");
13 continue;
14 }
15 for(i=0;i<m;++i) sg[i]=0;
16 for(i=m;i<=n;++i){
17 memset(vis,0,sizeof(vis));
18 for(j=1;j+m-1<=i;j++){
19 vis[sg[j-1]^sg[i-m-j+1]]=1;
20 }
21 for(j=0;;j++){
22 if(!vis[j]){
23 sg[i]=j;
24 break;
25 }
26 }
27 }
28 for(i=0;i<=n;++i){
29 if(sg[i]==0) sg[i]=1;
30 else sg[i]=0;
31 }
32 sg[n-m]?puts("aekdycoin"):puts("abcdxyzk");
33 }
34 return 0;
35 }
hdu-3980-nim博弈/sg函数
标签：mit java ini des nta ble href oev sea
原文地址：https://www.cnblogs.com/zzqc/p/9328414.html
