Arthur and his sister Caroll have been playing a game called Nim for some time now. Nim is played as follows:
recently learned an easy way to always be able to find the best move:
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.
- 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.
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.
- 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.
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
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.
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.
Output
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.
Print a newline after each test case.
Sample Input
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
Sample Output
LWW WWL
本来想着每组数据打一次表时间更短,没想到TLE了,哎,还是每个测试样例打一次表吧,直接按最大打表出来
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #define LL long long using namespace std; LL x,y; int n; int t; int l; int a[10010]; int f[110]; int sg[10010]; int vis[10010]; void getSG(int MAX) { memset(sg,0,sizeof(sg)); for(int i = 1; i < MAX; i++) { memset(vis,0,sizeof(vis)); for(int j = 0; f[j] <= i&&j<n; j++) { vis[sg[i-f[j]]] = 1; } for(int k = 0; ;k++) { if(!vis[k]) { sg[i] = k; break; } } } } void debug() { for(int i = 0; i < 20; i++) printf("%d\n",sg[i]); } int main() { while(scanf("%d",&n)!=EOF&&n) { for(int i = 0; i < n; i++) scanf("%d",&f[i]); sort(f,f+n); scanf("%d",&t); getSG(10010); while(t--) { scanf("%d",&l); int maxn = -1; for(int i = 0; i < l; i++) { scanf("%d",&a[i]); if(a[i]>maxn) maxn = a[i]; } //debug(); LL ans = 0; for(int i = 0; i < l; i++) { ans^=sg[a[i]]; } if(ans==0) printf("L"); else printf("W"); } printf("\n"); } return 0; }