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 = 100;

int sg[maxn + 10];
bool vis[maxn + 10];

int mex(){
    for (int i = 1; i <= maxn; i ++)
        if (not vis[i])
            return i;
}

int get_sg(int x){
    if (sg[x] != -1)
        return sg[x];

    memset(vis, false, sizeof vis);
    
    for (int i = 1; i < x; i ++){
        int temp = get_sg(i);
        vis[temp] = true;

        for (int j = 1; 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 % 8 == 0)
        return x - 1;
    if (x % 8 == 7)
        return x + 1;
    return x;
}

int main(){
#ifndef online_judge
    freopen(pron ".in", "r", stdin);
#endif
    int tcase, n, flag, temp;
    
    scanf("%d", &tcase);
    while (tcase --){
        flag = 0;

        scanf("%d", &n);
        for (int i = 0; i < n; i ++){
            scanf("%d", &temp);
            flag ^= get_sg(temp);
        }
            
        if (flag)
            puts("first player wins.");
        else
            puts("second player wins.");
    }
}

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

const int MAXN = 100;

bool vis[MAXN + 10];
int sg[MAXN + 10];

int mex(){
    for (int i = 1; i <= MAXN; i ++)
        if (not vis[i])
            return i;
}

int get_sg(int x){
    if (sg[x] != -1)
        return sg[x];

    memset(vis, false, sizeof vis);

    int temp;
    for (int i = 1; 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 % 4 == 0)
        return x - 1;
    if (x % 4 == 3)
        return x + 1;
    return x;
}

int main(){
#ifndef ONLINE_JUDGE
    freopen(PRON ".in", "r", stdin);
#endif
    memset(sg, -1, sizeof sg);

    int Tcase, n, temp, flag;

    scanf("%d", &Tcase);
    while (Tcase --){
        flag = 0;

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

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

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 = 100 + 10;
const int MAXN = 10000;

bool vis[MAXN + 10];
int k, Tcase, n, h, flag, s[MAXK], sg[MAXN + 10];

int mex(){
    for (int i = 0; i <= MAXN; i ++)
        if (not vis[i])
            return i;
}

void get_sg(){
    memset(sg, 0, sizeof sg);

    for (int i = 0, j; i <= MAXN; i ++){
        memset(vis, false, sizeof vis);

        j = 0;
        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) == 1 && k){
        for (int i = 0; i < k; i ++)
            scanf("%d", &s[i]);
        sort(s, s + k);

        get_sg();

        scanf("%d", &Tcase);
        while (Tcase --){
            flag = 0;
            
            scanf("%d", &n);
            for (int i = 0; i < n; i ++){
                scanf("%d", &h);
                flag ^= sg[h];
            }

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

        putchar(10);
    }
}

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 = 100;

int n, a[maxn];

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

    memset(a, 0, sizeof a);
    for (int i = 0, j = 0; i <= 60; i ++){
        while (j * (j + 1) / 2 <= i)
            ++ j;

        a[i] = -- j;
    }

    scanf("%d", &n);
    int flag = 0, p;
    while (n --){
        scanf("%d", &p);
        flag ^= a[p];
    }

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