Va 1378-A Funny Stone Game (combined Game)

Va 1378-A Funny Stone Game (combined Game)

Link to the Q & A 1378-A Funny Stone Game

Two people play the game. For a sequence, operations are in turn. The I, j, and k positions in each selected sequence require I

Solution: first, pre-process the SG value at each position. Then, for a given sequence, enumerate the position transition status and determine whether it is a mandatory or not.

#include 
  
   #include 
   
    #include using namespace std;const int maxn = 30;int n, g[maxn], s[maxn];int SG (int l) {    int vis[1000];    memset(vis, 0, sizeof(vis));    for (int i = 0; i < l; i++) {        for (int j = 0; j < l; j++)            vis[g[i]^g[j]] = 1;    }    int ret = -1;    while (vis[++ret]);    return ret;}void init () {    g[0] = 0;    for (int i = 1; i < maxn; i++)        g[i] = SG(i);}int judge () {    int ret = 0;    for (int i = 0; i < n-1; i++) {        if (s[i]&1)            ret ^= g[n-1-i];    }    return ret;}void put_ans () {    for (int i = 0; i < n-1; i++) {        if (s[i] == 0)            continue;        s[i]--;        for (int j = i+1; j < n; j++) {            s[j]++;            for (int k = j; k < n; k++) {                s[k]++;                if (judge() == 0) {                    printf(" %d %d %d\n", i, j, k);                    return;                }                s[k]--;            }            s[j]--;        }        s[i]++;    }}int main () {    init();    int cas = 1;    while (scanf("%d", &n) == 1 && n) {        for (int i = 0; i < n; i++)            scanf("%d", &s[i]);        printf("Game %d:", cas++);        if (judge())            put_ans();        else            printf(" -1 -1 -1\n");    }    return 0;}