HDU 1536 S-Nim
2013-07-17 22:53
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S-Nim
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3178 Accepted Submission(s): 1406
[align=left]Problem Description[/align]
Arthur and his sister Caroll have been playing a game called Nim for some time now. Nim is played as follows:
The starting position has a number of heaps, all containing some, not necessarily equal, number of beads.
The players take turns chosing a heap and removing a positive number of beads from it.
The first player not able to make a move, loses.
Arthur and Caroll really enjoyed playing this simple game until they recently learned an easy way to always be able to find the best move:
Xor the number of beads in the heaps in the current position (i.e. if we have 2, 4 and 7 the xor-sum will be 1 as 2 xor 4 xor 7 = 1).
If the xor-sum is 0, too bad, you will lose.
Otherwise, move such that the xor-sum becomes 0. This is always possible.
It is quite easy to convince oneself that this works. Consider these facts:
The player that takes the last bead wins.
After the winning player's last move the xor-sum will be 0.
The xor-sum will change after every move.
Which means that if you make sure that the xor-sum always is 0 when you have made your move, your opponent will never be able to win, and, thus, you will win.
Understandibly it is no fun to play a game when both players know how to play perfectly (ignorance is bliss). Fourtunately, Arthur and Caroll soon came up with a similar game, S-Nim, that seemed to solve this problem. Each player is now only allowed to remove
a number of beads in some predefined set S, e.g. if we have S =(2, 5) each player is only allowed to remove 2 or 5 beads. Now it is not always possible to make the xor-sum 0 and, thus, the strategy above is useless. Or is it?
your job is to write a program that determines if a position of S-Nim is a losing or a winning position. A position is a winning position if there is at least one move to a losing position. A position is a losing position if there are no moves to a losing position.
This means, as expected, that a position with no legal moves is a losing position.
[align=left]Input[/align]
Input consists of a number of test cases. For each test case: The first line contains a number k (0 < k ≤ 100 describing the size of S, followed by k numbers si (0 < si ≤ 10000) describing S. The second line contains a number m (0
< m ≤ 100) describing the number of positions to evaluate. The next m lines each contain a number l (0 < l ≤ 100) describing the number of heaps and l numbers hi (0 ≤ hi ≤ 10000) describing the number of beads in the heaps. The last test case is followed by
a 0 on a line of its own.
[align=left]Output[/align]
For each position: If the described position is a winning position print a 'W'.If the described position is a losing position print an 'L'. Print a newline after each test case.
[align=left]Sample Input[/align]
2 2 5
3
2 5 12
3 2 4 7
4 2 3 7 12
5 1 2 3 4 5
3
2 5 12
3 2 4 7
4 2 3 7 12
0
[align=left]Sample Output[/align]
LWW
WWL
[align=left]Source[/align]
Norgesmesterskapet 2004
[align=left]Recommend[/align]
LL
最基础的运用SG函数
贴两份代码仅供参考
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; int sg[11000],k,st[110],vis[110]; void getsg() { int i,j; sg[0]=0; for(i=1;i<=10000;i++) { memset(vis,0,sizeof(vis)); for(j=0;j<k;j++) { if(i < st[j])break; if(sg[i-st[j]]<101) vis[sg[i-st[j]]]=1; } for(j=0;;j++) if(!vis[j]) break; sg[i]=j; } } int main() { int m,n,i,v,sum; while(scanf("%d",&k),k) { for(i=0;i<k;i++) scanf("%d",&st[i]); sort(st,st+k); scanf("%d",&m); getsg(); while(m--) { sum=0; scanf("%d",&n); for(i=0;i<n;i++) { scanf("%d",&v); sum^=sg[v]; } putchar(sum?'W':'L'); } putchar('\n'); } return 0; }
#include<stdio.h> #include<string.h> #include<iostream> #include<string> #include <algorithm> using namespace std; #define N 10010 int s ; int GS ; int gs(int x, int k) { int mex[101]; memset(mex,0,sizeof(mex)); if(GS[x] != -1) return GS[x]; if(x - s[0] < 0) return GS[x] = 0; for(int i = 0; i < k && x-s[i] >= 0; i++) mex[gs(x-s[i], k)] = 1; for(int i = 0;;i++) if(!mex[i]) return GS[x] = i; } int main () { int k, i, j; int T, n, a; while(scanf("%d", &k),k) { memset(GS,-1,sizeof(GS)); GS[0] = 0; for(i = 0; i < k; i++) scanf("%d", &s[i]); //可移动石子的个数 sort(s,s+k); string ans; scanf("%d", &T); while(T--) { int temp = 0; scanf("%d", &n); for(i = 0; i < n; i++) { scanf("%d", &a); temp ^= gs(a,k); } if(!temp) ans += "L"; else ans += "W"; } cout << ans << endl; } return 0; }
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