同余问题，乘法模逆元【模板】

扩展欧几里得，求一组解x，y，使得gcd(a，b)  = d = a * x + b * y

void ExGcd(int a,int b,int &d,int &x,int &y)
{
    if(b == 0)
    {
        x = 1;
        y = 0;
        d = a;
    }
    else
    {
        ExGcd(b,a%b,d,y,x);
        y -= x*(a/b);
    }
}

扩展欧几里得，求所有解x，y，使得c = a * x + b * y

bool ModeEqual(int a,int b,int c)   //解a*x + b*y = c;a*x ≡ c(mod b)
{
    int x,y,d,x0;
    ExGcd(a,b,d,x,y);
    if(c%d) return false;   //无解
    x0 = x * (c/d) % b;
    for(int i = 1; i < d; ++i)  //输出所有解
        printf("%d\n",(x0+i*(b/d))%b);
    return true;
}

扩展欧几里得，求a关于n的逆元a^-1，使得a * a^-1 ≡ 1(mod n)

int ModInverse(int a,int n,int &d,int &x,int &y)
{
    ExGcd(a,n,d,x,y);
	if(1 % d != 0)
		return -1;  //不存在模逆元
	int ans = x/d < 0 ? x/d + n : x/d;
	return ans;     //返回模逆元a^-1
}

扩展欧几里得，求解x，满足同余方程组x ≡ Ri(mod Ai)

int ModEquals(int N)  //解方程组x ≡ Ri(mod Ai)
{
    int a,b,d,x,y,c,A1,R1,A2,R2;
    bool flag = 1;  //标记是否有解
    scanf("%d%d",&A1,&R1);
    for(int i = 1; i < N; ++i)
    {
        scanf("%d%d",&A2,&R2);
        a = A1, b = A2, c = R2 - R1;
        ExGcd(a,b,d,x,y);
        if(c % d != 0)
            flag = 0;
        int t = b/d;
        x = (x*(c/d)%t + t) % t;
        R1 = A1 * x + R1;
        A1 = A1 * (A2 / d);
    }
    if( !flag )  R1 = -1;
    return R1;  //求出解，-1表示无解
}

扩展欧几里得，求解x，满足高次同余方程A^x ≡ B(mod C)

#define LL __int64
const int MAXN = 65535;

struct HASH
{
    int a;
    int b;
    int next;
}Hash[MAXN*2];

int flag[MAXN+66];
int top,idx;

void ins(int a,int b)
{
    int k = b & MAXN;
    if(flag[k] != idx)
    {
        flag[k] = idx;
        Hash[k].next = -1;
        Hash[k].a = a;
        Hash[k].b = b;
        return;
    }
    while(Hash[k].next != -1)
    {
        if(Hash[k].b == b)
            return;
        k = Hash[k].next;
    }
    Hash[k].next = ++top;
    Hash[top].next = -1;
    Hash[top].a = a;
    Hash[top].b = b;
}

int Find(int b)
{
    int k = b & MAXN;
    if(flag[k] != idx)
        return -1;
    while(k != -1)
    {
        if(Hash[k].b == b)
            return Hash[k].a;
        k = Hash[k].next;
    }
    return -1;
}

int GCD(int a,int b)
{
    if(b == 0)
        return a;
    return GCD(b,a%b);
}

void ExGcd(int a,int b,int &d,int &x,int &y)
{
    if(b == 0)
    {
        x = 1;
        y = 0;
        d = a;
    }
    else
    {
        ExGcd(b,a%b,d,y,x);
        y -= x*(a/b);
    }
}

int Inval(int a,int b,int n)
{
    int x,y,d,e;
    ExGcd(a,n,d,x,y);
    e = (LL)x*b%n;
    return e < 0 ? e + n : e;
}

int PowMod(LL a,int b,int c)
{
    LL ret = 1%c;
    a %= c;
    while(b)
    {
        if(b&1)
            ret = ret*a%c;
        a = a*a%c;
        b >>= 1;
    }
    return ret;
}

int BabyStep(int A,int B,int C) //解A^x ≡ B(mod C)
{
    top = MAXN;
    ++idx;
    LL buf = 1%C,D = buf,K;
    int d = 0,temp,i;
    for(i = 0; i <= 100; buf = buf*A%C,++i)
    {
        if(buf == B)
            return i;
    }
    while((temp = GCD(A,C)) != 1)
    {
        if(B % temp)
            return -1;
        ++d;
        C /= temp;
        B /= temp;
        D = D*A/temp%C;
    }
    int M = (int)ceil(sqrt((double)C));
    for(buf = 1%C,i = 0; i <= M; buf = buf*A%C,++i)
        ins(i,buf);
    for(i = 0,K = PowMod((LL)A,M,C); i <= M; D = D*K%C,++i)
    {
        temp = Inval((int)D,B,C);
        int w;
        if(temp >= 0 && (w = Find(temp)) != -1)
            return i * M + w + d;
    }
    return -1;  //无解
}