HDU Game Theory

HDU 5795 || 3032

把x个石子的堆分成非空两(i, j)或三堆(i, j, k)的操作->(sg[i] ^ sg[j])或(sg[i] ^ sg[j] ^ sg[k])是x的后继

 #define pron "hdu5795"
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 const int maxn = ;

 int sg[maxn + ];
 bool vis[maxn + ];

 int mex(){
     for (int i = ; i <= maxn; i ++)
         if (not vis[i])
             return i;
 }

 int get_sg(int x){
     if (sg[x] != -)
         return sg[x];

     memset(vis, false, sizeof vis);

     for (int i = ; i < x; i ++){
         int temp = get_sg(i);
         vis[temp] = true;

         for (int j = ; j < x - i; j ++)
             vis[temp ^ get_sg(j) ^ get_sg(x - i - j)] = true;
     }

     return sg[x] = mex();
 }

 int get_sg(int x){
     if (x %  == )
         return x - ;
     if (x %  == )
         return x + ;
     return x;
 }

 int main(){
 #ifndef online_judge
     freopen(pron ".in", "r", stdin);
 #endif
     int tcase, n, flag, temp;

     scanf("%d", &tcase);
     while (tcase --){
         flag = ;

         scanf("%d", &n);
         for (int i = ; i < n; i ++){
             scanf("%d", &temp);
             flag ^= get_sg(temp);
         }

         if (flag)
             puts("first player wins.");
         else
             puts("second player wins.");
     }
 }

5795

 #define PRON "hdu3032"
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 const int MAXN = ;

 bool vis[MAXN + ];
 int sg[MAXN + ];

 int mex(){
     for (int i = ; i <= MAXN; i ++)
         if (not vis[i])
             return i;
 }

 int get_sg(int x){
     if (sg[x] != -)
         return sg[x];

     memset(vis, false, sizeof vis);

     int temp;
     for (int i = ; i < x; i ++){
         temp = get_sg(i);
         vis[temp] = vis[temp ^ get_sg(x - i)] = true;
     }

     return sg[x] = mex();
 }

 int get_SG(int x){
     if (x %  == )
         return x - ;
     if (x %  == )
         return x + ;
     return x;
 }

 int main(){
 #ifndef ONLINE_JUDGE
     freopen(PRON ".in", "r", stdin);
 #endif
     memset(sg, -, sizeof sg);

     int Tcase, n, temp, flag;

     scanf("%d", &Tcase);
     while (Tcase --){
         flag = ;

         scanf("%d", &n);
         while (n --){
             scanf("%d", &temp);
             flag ^= get_SG(temp);
         }

         if (flag)
             puts("Alice");
         else
             puts("Bob");
     }
 }

3032

HDU 1536 || 1944

直接get_sg，不需要打表找规律

 #define PRON "hdu1536"
 #include <cstdio>
 #include <cstring>
 #include <iostream>
 #include <algorithm>
 using namespace std;

 const int MAXK =  + ;
 const int MAXN = ;

 bool vis[MAXN + ];
 int k, Tcase, n, h, flag, s[MAXK], sg[MAXN + ];

 int mex(){
     for (int i = ; i <= MAXN; i ++)
         if (not vis[i])
             return i;
 }

 void get_sg(){
     memset(sg, , sizeof sg);

     for (int i = , j; i <= MAXN; i ++){
         memset(vis, false, sizeof vis);

         j = ;
         while (i >= s[j] && j < k)
             vis[sg[i - s[j ++]]] = true;

         sg[i] = mex();
     }
 }

 int main(){
 #ifndef ONLINE_JUDGE
     freopen(PRON ".in", "r", stdin);
 #endif

     while (scanf("%d", &k) ==  && k){
         for (int i = ; i < k; i ++)
             scanf("%d", &s[i]);
         sort(s, s + k);

         get_sg();

         scanf("%d", &Tcase);
         while (Tcase --){
             flag = ;

             scanf("%d", &n);
             for (int i = ; i < n; i ++){
                 scanf("%d", &h);
                 flag ^= sg[h];
             }

             if (flag)
                 putchar('W');
             else
                 putchar('L');
         }

         putchar();
     }
 }

1536

Codeforces 768E

取石子，但是对于每一堆，不能取相同的个数。比如这一堆被拿走了x个，下一次就不能再拿x个了。

官方题解用的dp。但这道题可以...

处理一个数组a[i], 表示取完i个石子最多需要的步骤数，即 1 + 2 + .... + a[i] <= i。由于是后手，所以前者采取什么策略我就可以采取什么策略，因此求一个异或和就好。

#define PRON "768e"
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;

const int maxn = ;

int n, a[maxn];

int main(){
#ifndef ONLINE_JUDGE
    freopen(PRON ".in", "r", stdin);
#endif

    memset(a, , sizeof a);
    for (int i = , j = ; i <= ; i ++){
        while (j * (j + ) /  <= i)
            ++ j;

        a[i] = -- j;
    }

    scanf("%d", &n);
    int flag = , p;
    while (n --){
        scanf("%d", &p);
        flag ^= a[p];
    }

    puts(flag ? "NO" : "YES");
}

768E