HDU 1535 S-Nim(SG函数)
S-Nim
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8729 Accepted Submission(s): 3660
Problem Description
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.
Input
Output
Sample Input
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
Sample Output
题意
首先给出k,表示有几种每次取石子个数的集合,即给出123,每次可以取1,2,3个。然后给出询问次数m。每次询问给出n堆石子,然后询问当前状态是P还是N。
分析
SG函数。两种求法,都写了一遍。耗时间差不多
code
预处理
#include<cstdio>
#include<cstring>
#include<algorithm> using namespace std; const int N = ;
int sg[],f[];
int k,n,m;
bool Hash[]; void get_SG() {
memset(sg,,sizeof(sg));
for (int i=; i<=N; ++i) {
memset(Hash,false,sizeof(Hash));
for (int j=; j<=k&&f[j]<=i; ++j)
Hash[sg[i-f[j]]] = true;
for (int j=; j<=N; ++j)
if (!Hash[j]) {sg[i] = j;break;}
}
} int main () {
while (~scanf("%d",&k) && k) {
for (int i=; i<=k; ++i)
scanf("%d",&f[i]);
sort(f+,f+k+);
get_SG();
scanf("%d",&m);
for (int i=; i<=m; ++i) {
scanf("%d",&n);
int ans = ;
for (int t,j=; j<=n; ++j) {
scanf("%d",&t);
ans ^= sg[t];
}
if (ans == ) printf("L");
else printf("W");
}
puts("");
}
return ;
}
dfs记忆化搜索
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std; int sg[],f[];
int k,n,m; int get_SG(int x) {
if (sg[x] != -) return sg[x];
bool Hash[]; //不能再外面开
memset(Hash,false,sizeof(Hash));
for (int i=; i<=k; ++i) {
if (f[i] > x) break;
get_SG(x-f[i]);
Hash[sg[x-f[i]]] = true;
}
for (int i=; ; ++i)
if (!Hash[i]) {sg[x] = i;break;}
return sg[x];
}
int main () {
while (~scanf("%d",&k) && k) {
for (int i=; i<=k; ++i)
scanf("%d",&f[i]);
sort(f+,f+k+);
memset(sg,-,sizeof(sg));
sg[] = ;
scanf("%d",&m);
for (int i=; i<=m; ++i) {
scanf("%d",&n);
int ans = ;
for (int t,j=; j<=n; ++j) {
scanf("%d",&t);
ans ^= get_SG(t);
}
if (ans == ) printf("L");
else printf("W");
}
puts("");
}
return ;
}
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