Baby-Step Giant-Step 解高次同余方程（形如(a^x)%n=b,其中n为素数）

//baby_step giant_step// a^x = b (mod n) n为素数，a,b < n// 求解上式 0<=x < n的解#define MOD 76543int hs[MOD],head[MOD],next[MOD],id[MOD],top;void insert(int x,int y) {    int k = x%MOD;    hs[top] = x, id[top] = y, next[top] = head[k], head[k] = top++;}int find(int x) {    int k = x%MOD;    for(int i = head[k]; i != -1; i = next[i])        if(hs[i] == x)            return id[i];    return -1;}int BSGS(int a,int b,int n) {    memset(head,-1,sizeof(head));    top = 1;    if(b == 1)return 0;    int m = sqrt(n*1.0), j;    long long x = 1, p = 1;    for(int i = 0; i < m; ++i, p = p*a%n)insert(p*b%n,i);    for(long long i = m; ; i += m) {        if( (j = find(x = x*p%n)) != -1 )return i-j;        if(i > n)break;    }    return -1;}