Congruence problem, multiplication modulo inverse "template"

Expand Euclid, find a group of x, Y, making 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);    }}

Expand Euclidean, ask for all solution x, y, make C = a * x + b * y

BOOL Modeequal (int a,int b,int c)   //Solution 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;   No solution    x0 = x * (C/D)% B;    for (int i = 1; i < D; ++i)  //Output All        of the solutions printf ("%d\n", (x0+i* (b/d))%b);    return true;}

Expand Euclid, ask a about the inverse of n a^-1, make 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;  Non-existent modulo inverse int ans = x/d < 0? X/d + n:x/d;return ans;     return modulo inverse a^-1}

Extended Euclidean, Solution X, satisfies the congruence equation set X≡ri (mod Ai)

int modequals (int N)  //Solution equation set X≡ri (mod Ai) {    int a,b,d,x,y,c,a1,r1,a2,r2;    BOOL flag = 1;  Mark whether there is a solution    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;  Find solution, 1 means no solution}

Extended Euclidean, Solution X, satisfies high-time congruence equation a^x≡b (mod C)

#define LL __int64const 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)//Solution a^x≡b (mod C) {top = MAXN;    ++idx;    LL buf = 1%c,d = Buf,k;    int d = 0,temp,i;    for (i = 0; I <=; 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; No solution} 

Congruence problem, multiplication modulo inverse "template"