SG博弈简单题

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ZOJ - 2083 - Win the Game

题目传送：Win the Game

最近正在慢慢体会博弈里面的SG函数的意义

此题是最简单的SG博弈问题，只需打个表就OK了

AC代码：

#include <map>#include <set>#include <list>#include <cmath>#include <deque>#include <queue>#include <stack>#include <bitset>#include <cctype>#include <cstdio>#include <string>#include <vector>#include <complex>#include <cstdlib>#include <cstring>#include <fstream>#include <sstream>#include <utility>#include <iostream>#include <algorithm>#include <functional>#define LL long long#define INF 0x7fffffffusing namespace std;int sg[55];int vis[55];void get_sg() {    sg[0] = 0;    sg[1] = 0;    for(int i = 2; i < 55; i ++) {        memset(vis, 0, sizeof(vis));        vis[sg[i - 2]] = 1;        for(int j = 1; j + 2 <= i; j ++) {            vis[sg[j] ^ sg[i - j - 2]] = 1;        }        for(int j = 0;; j ++) {            if(!vis[j]) {                sg[i] = j;                break;            }        }    }}int n;int main() {    get_sg();    //for(int i = 0; i < 55; i ++) cout << sg[i] << " ";    while(scanf("%d", &n) != EOF) {        int SG = 0, t;        for(int i = 0; i < n; i ++) {            scanf("%d", &t);            SG ^= sg[t];        }        if(SG == 0) {            printf("No\n");        }        else printf("Yes\n");    }    return 0;}

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ZOJ - 2507 - Let’s play a game

题目传送：Let’s play a game

AC代码：

#include <map>#include <set>#include <list>#include <cmath>#include <deque>#include <queue>#include <stack>#include <bitset>#include <cctype>#include <cstdio>#include <string>#include <vector>#include <complex>#include <cstdlib>#include <cstring>#include <fstream>#include <sstream>#include <utility>#include <iostream>#include <algorithm>#include <functional>#define LL long long#define INF 0x7fffffffusing namespace std;int T;int main() {    scanf("%d", &T);    while(T --) {        int n, t, cnt = 0;;        int SG = 0;        scanf("%d", &n);        for(int i = 0; i < n; i ++) {            scanf("%d", &t);            SG ^= t;            if(t > 1) cnt ++;        }        if(cnt == 0) {            if(SG == 1) {                printf("2\n");            }            else printf("1\n");        }        else {            if(SG) {                printf("1\n");            }            else printf("2\n");        }    }    return 0;}

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ZOJ - 3529 - A Game Between Alice and Bob

题目传送：A Game Between Alice and Bob

和之前UVA那个题类似，也是打表筛出素因子个数，然后套SG即可

AC代码：

#include <map>#include <set>#include <list>#include <cmath>#include <deque>#include <queue>#include <stack>#include <bitset>#include <cctype>#include <cstdio>#include <string>#include <vector>#include <complex>#include <cstdlib>#include <cstring>#include <fstream>#include <sstream>#include <utility>#include <iostream>#include <algorithm>#include <functional>#define LL long long#define INF 0x7fffffffusing namespace std;const int maxn = 5000005;int n;int a[100005];int chu[maxn];int to[maxn];void init() {    to[1] = 0;    for(int i = 2; i < maxn; i ++) chu[i] = i;    for(int i = 2; i < maxn; i ++) {        if(to[i]) continue;        to[i] ++;        for(int j = 2 * i; j < maxn; j += i) {            while(chu[j] % i == 0) {                to[j] ++;                chu[j] /= i;            }        }    }}int main() {    init();    //for(int i = 1; i < 10; i ++) cout << to[i] << " ";    int cas = 1;    while(scanf("%d", &n) != EOF) {        int SG = 0, t;        for(int i = 0; i < n; i ++) {            scanf("%d", &t);            a[i] = to[t];                   SG ^= a[i];        }        if(SG == 0) {            printf("Test #%d: Bob\n", cas);        }        else {            printf("Test #%d: Alice", cas);            for(int i = 0; i < n; i ++) {                if(a[i] > (a[i] ^ SG)) {                    printf(" %d\n", i + 1);                    break;                }            }        }        cas ++;    }    return 0;}

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ZOJ - 3513 - Human or Pig

题目传送：Human or Pig

注意题目要输出的P点和H点的含义。

含义是：

• P点，此时的状态为猪，且随便怎么走都能够满足条件，即落入水中时的状态为猪，然后可以永远变为人类
• H点，此时的状态为人类，且只需要下一状态有一个为猪即可，因为状态为猪时随便走都可以满足条件

AC代码：

#include <map>#include <set>#include <list>#include <cmath>#include <deque>#include <queue>#include <stack>#include <bitset>#include <cctype>#include <cstdio>#include <string>#include <vector>#include <complex>#include <cstdlib>#include <cstring>#include <fstream>#include <sstream>#include <utility>#include <iostream>#include <algorithm>#include <functional>#define LL long long#define INF 0x7fffffffusing namespace std;const int maxn = 40005;int mp[maxn];//由于X*Y<=40000，但是这里开40000*40000的数组又太大了，所以这里我们应该用一维数组来模拟二维数组int X, Y;//行数以及列数int& get_pos(int x, int y) {//获取一维数组上对应的二维数组上的那个值    return mp[(x - 1) * Y + y];}//这里需要注意P点和H点的含义，即//H点代表此时只能是人类，才能确保落入水中时的状态是猪//P点代表此时是猪，也能确保落入水中时的状态是猪，这一步只需要保证他的后继状态都是H或者直接掉入水中即可////通俗一点的来说就是随便乱走都可以满足条件的就是P点，然后可以根据后继状态有P来确保满足条件的就是H点char get_char(int x, int y) {//获取一次的    int &ret = get_pos(x, y);    ret = 'P';//先初始化为P点，P点代表此时是猪，因为猪是没思想的，可以随便走，所以初始化为这个    //只要当前状态的后继状态出现了P点就可以转化成H点，因为H点代表人类，人类是有思想的，可以指定选择走到P点，而走到P点后随意走都可以满足条件    //如果当前状态的后继状态出现了P点，而又没有转化，则当前不能确保能满足条件    for(int k = 1; y > k * x; k ++) {        if(get_pos(x, y - k * x) == 'P') {            ret = 'H';            break;        }    }    for(int k = 1; x > k * y; k ++) {        if(get_pos(x - k * y, y) == 'P') {            ret = 'H';            break;        }    }    return ret;}int main() {    int cas = 1;    while(scanf("%d %d", &X, &Y) != EOF) {        printf("Case #%d:\n", cas ++);        for(int i = 1; i <= X; i ++) {            for(int j = 1; j <= Y; j ++) {                putchar(get_char(i, j));            }            puts("");        }    }    return 0;}

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