HDU1536-S-Nim(SG函数和异或处理)
题目链接 S-Nim
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9613 Accepted Submission(s): 3967
Problem Description
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.
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.
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.
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 3 712
0
Sample Output
LWW
WWL
Source
Norgesmesterskapet 2004
题目意思：就是首先给你一个数s,输入s个数的集合，接下来的操作只能取与这集合相对应的石子，接下来一行输入一个数m,表示
m次操作，然后输入一个n,表示有n堆石子，如果先手胜出输出L,否则输出W,当然得一连串输出这m个字母。
思路:运用sg函数和xor处理，就是一个裸题了。
tle代码:这个原始得模板比较low，容易超时。后来发现超时是因为访问数组的问题，如果设置为bool就不超时了。
#include
#include
#include
#include
using namespace std;
const int maxn=10010;
int sg[maxn],oper[maxn],temp[maxn],k[maxn];
int s;
void getsg()
{
memset(sg,0,sizeof(sg));
for(int i=0;i<10010;i++){
memset(temp,0,sizeof(temp));
for(int j=0;j 参考的好用板子:https://blog.csdn.net/qq_40932661/article/details/81308957
AC代码①:
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int maxn = 10010;
int sg[maxn], s[maxn],k;//sg函数值 s是可取的数（1，3，4）
bool vis[maxn];//标记
void SG(int n) {
memset(sg, 0, sizeof(sg));
for(int i = 1; i < maxn; i++) {
memset(vis, false, sizeof(vis));
for(int j = 0; j < n && s[j] <= i; j++)//s[j] <= i 表示后继 j < n 共有n种取法
vis[sg[i-s[j]]] = true;//把后继的sg值标记，出现过了
for(int j = 0; j <= maxn; j++) {
if(!vis[j]) {//第一个没有出现的数字 sort优化
sg[i] = j;
break;
}
}
}
}
int main() {
int m, l, x;
while(scanf("%d", &k),k) {
for(int i = 0; i < k; i++)
scanf("%d", &s[i]);
sort(s, s+k);//优化
SG(k);
scanf("%d", &m);
while(m--) {
scanf("%d", &l);
int sum = 0;
while(l--) {
scanf("%d", &x);
sum ^= sg[x];//每一堆的异或和
}
if(sum == 0)//必败 P
printf("L");
else// 必胜 N
printf("W");
}
printf("\n");
}
} AC代码②:
#include
#include
#include
#include
#include
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
int sg[10010], s[110],k;//意义同上
int sG(int x) {
if(sg[x] != -1) //已经计算过直接返回
return sg[x];
bool vis[110];
memset(vis, 0, sizeof(vis));//标记
for(int i = 0; i < k; i++) {//跑可取数
if(x >= s[i]) {//如果是后继
sG(x-s[i]);//递归得到后继的sg值
vis[sg[x-s[i]]] = 1;//把后继的sg值标记出现过
}
}
int e;
for(int i = 0;;i++) {//第一个没有出现的
if(!vis[i]) {
e = i;
break;
}
}
return sg[x] = e;//返回值
}
int main() {
int m, l, x;
while(~scanf("%d", &k)&&k) {
for(int i = 0; i < k; i++)
scanf("%d", &s[i]);
memset(sg, -1, sizeof(sg));//初始化-1
sort(s, s+k);//排序
scanf("%d", &m);
while(m--) {
scanf("%d", &l);
int sum = 0;
while(l--) {
scanf("%d", &x);
sum ^= sG(x);
}
if(sum == 0)
printf("L");
else
printf("W");
}
printf("\n");
}
}