Ultraviolet A 10561-Treblecross (Nim)

Ultraviolet A 10561-Treblecross (Nim)

Link to the question: Ultraviolet A 10561-Treblecross

N grids are arranged in a row, some of which are 'x'. The two players take turns and put X in the grid. If three X boxes appear in a row, they will win. Shows whether the first hand can win the game. The first step is how to proceed.

Solution: one X can cause two grids on either side to no longer be placed on X, because if XX .. XX,. XX, and X. X appear, the next person will surely win. Therefore, for the lattice sequence whose length is n, g (x) = maxg (x? 3), g (x? 4)... g (2) XORg (n? 7 ).
Therefore, you can process the g array in advance, and then enumerate the positions under the first hand for the given sequence to determine whether the latter hand is defeated in this case. Note: If you can win the game first, you must consider it.

 #include 
  
   #include 
   
    #include using namespace std;const int maxn = 200;int g[maxn+5], v[maxn+5];int SG (int s) {    memset(v, 0, sizeof(v));    for (int i = 1; i <= s; i++) {        int t = 0;        if (s - i - 2 >= 0)            t ^= g[s-i-2];        if (i - 3 >= 0)            t ^= g[i-3];        v[t] = 1;    }    int mv = -1;    while (v[++mv]);    return mv;}void init () {    memset(g, 0, sizeof(g));    g[1] = g[2] = g[3] = 1;    for (int i = 4; i <= 200; i++)        g[i] = SG(i);}int n, len, pos[maxn+5];char str[maxn+5];bool check () {    int pre = -3, ret = 0;    for (int i = 0; i <= len; i++) {        if (str[i] == 'X') {            if (i - pre <= 2)                return false;            int l = max(i - pre - 5, 0);            if (l)                ret ^= g[l];            pre = i;        }    }    int l = max(len - pre - 3, 0);    if (l)        ret ^= g[l];    return ret == 0;}bool judge () {    n = 0;    len = strlen(str);    for (int i = 0; i < len; i++) {        if (str[i] == '.') {            str[i] = 'X';            if (check())                pos[n++] = i + 1;            str[i] = '.';        }        if (i && i < len - 1 & str[i-1] == 'X' && str[i+1] == 'X')                pos[n++] = i + 1;        if (i < len - 2 && str[i+1] == 'X' && str[i+2] == 'X')            pos[n++] = i + 1;        if (i >= 2 && str[i-1] == 'X' && str[i-2] == 'X')            pos[n++] = i + 1;    }    return n;}int main () {    init();    int cas;    scanf("%d", &cas);    while (cas--) {        scanf("%s", str);        printf("%s\n", judge() ? "WINNING" : "LOSING");        if (n)            printf("%d", pos[0]);        for (int i = 1; i < n; i++)            printf(" %d", pos[i]);        printf("\n");    }    return 0;}