H - S-Nim
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 rst 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 7 12 0
Sample Output
LWW WWL
题目大意：
给定 K个数，表示对应的操作集的长度
接着是操作集合 S
再给定 m次要评估的游戏
每次游戏都会给定 堆的个数 L
然后给定一个堆集合 heap
思路：
求出每个堆的SG ，再对每个堆进行异或运算，若为0 则，会失败。反之胜利。
具体思路见 ：https://blog.csdn.net/qq_40507857/article/details/81277606
#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
//heap number
const int MAXN = 10010;
//k number
const int MAXK = 110;
bool p[MAXN];//标记
int calSG(int *S,int Snum,int max,int *SG,int maxHeap){
//bool *p = new bool[max+1]; // time limit this way
//fill(p,p+max+1,false); // time limit this way
memset(p,false,sizeof(p));
for(int i=0;i<maxHeap;i++){
for(int j=0;j<Snum&&i>=S[j];j++){
p[SG[i-S[j]]] = true;
}
for(int k=0;k<=max;k++){
if(p[k]==false){
SG[i] = k;
break;
}
}
memset(p,false,sizeof(p));
}
return 0;
}
int main(){
int S[MAXK];
int heap[MAXK];
int SG[MAXN];
int Snum;
while(scanf("%d",&Snum)!=EOF){
if(Snum==0) break;
int max=-1;
for(int i=0;i<Snum;i++){
scanf("%d",&S[i]);
if(max<S[i]){
max = S[i];
}
}
sort(S,S+Snum);
calSG(S,Snum,max,SG,MAXN);
int M;
scanf("%d",&M);
for(int i=0;i<M;i++){
int L;
scanf("%d",&L);
int sg =0;
for(int j=0;j<L;j++){
scanf("%d",&heap[j]);
sg=sg^SG[heap[j]];
}
if(sg==0){
printf("L");
}else{
printf("W");
}
}
printf("\n");
}
return 0;
}
注意：
//bool *p = new bool[max+1]; // time limit this way
//fill(p,p+max+1,false); // time limit this way
在方法中用动态申请地址的方法会超时，所以还是再最开始申请静态并结合memset(p,false,sizeof(p));的方法会减少时间复杂度