POJ 3243 Clever Y BSGS 算法 （模板）

Clever Y
Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 8920 Accepted: 2222
Description

Little Y finds there is a very interesting formula in mathematics:

XY mod Z = K

Given X, Y, Z, we all know how to figure out K fast. However, given X, Z, K, could you figure out Y fast?

Input

Input data consists of no more than 20 test cases. For each test case, there would be only one line containing 3 integers X, Z, K (0 ≤ X, Z, K ≤ 109).
Input file ends with 3 zeros separated by spaces.
Output

For each test case output one line. Write “No Solution” (without quotes) if you cannot find a feasible Y (0 ≤ Y < Z). Otherwise output the minimum Y you find.
Sample Input

5 58 33
2 4 3
0 0 0
Sample Output

9
No Solution

BSGS 算法主要解决的是A^x=B （mod C）这类高次同余方程 的问题

证明的话 大概就是把x替换一下 令x=i*k-j的形式 然后预处理一部分放到hash表中 然后在枚举另外一部分，看能不能在hash表中找到。

#include<cstdio>#include<cstring>#include<cmath>using namespace std;#define lint __int64#define MAXN 131071struct HashNode { lint data, id, next; };HashNode hash[MAXN<<1];bool flag[MAXN<<1];lint top;void Insert ( lint a, lint b ){    lint k = b & MAXN;    if ( flag[k] == false )    {        flag[k] = true;        hash[k].next = -1;        hash[k].id = a;        hash[k].data = b;        return;    }    while( hash[k].next != -1 )    {        if( hash[k].data == b ) return;        k = hash[k].next;    }    if ( hash[k].data == b ) return;    hash[k].next = ++top;    hash[top].next = -1;    hash[top].id = a;    hash[top].data = b;}lint Find ( lint b ){    lint k = b & MAXN;    if( flag[k] == false ) return -1;    while ( k != -1 )    {        if( hash[k].data == b ) return hash[k].id;        k = hash[k].next;    }    return -1;}lint gcd ( lint a, lint b ){    return b ? gcd ( b, a % b ) : a;}lint ext_gcd (lint a, lint b, lint& x, lint& y ){    lint t, ret;    if ( b == 0 )    {        x = 1, y = 0;        return a;    }    ret = ext_gcd ( b, a % b, x, y );    t = x, x = y, y = t - a / b * y;    return ret;}lint mod_exp ( lint a, lint b, lint n ){    lint ret = 1;    a = a % n;    while ( b >= 1 )    {        if( b & 1 )            ret = ret * a % n;        a = a * a % n;        b >>= 1;    }    return ret;}lint BabyStep_GiantStep ( lint A, lint B, lint C ){    top = MAXN;  B %= C;    lint tmp = 1, i;    for ( i = 0; i <= 100; tmp = tmp * A % C, i++ )        if ( tmp == B % C ) return i;    lint D = 1, cnt = 0;    while( (tmp = gcd(A,C)) !=1 )    {        if( B % tmp ) return -1;        C /= tmp;        B /= tmp;        D = D * A / tmp % C;        cnt++;    }    lint M = (lint)ceil(sqrt(C+0.0));    for ( tmp = 1, i = 0; i <= M; tmp = tmp * A % C, i++ )        Insert ( i, tmp );    lint x, y, K = mod_exp( A, M, C );    for ( i = 0; i <= M; i++ )    {        ext_gcd ( D, C, x, y ); // D * X = 1 ( mod C )        tmp = ((B * x) % C + C) % C;        if( (y = Find(tmp)) != -1 )            return i * M + y + cnt;        D = D * K % C;    }    return -1;}int main(){    lint A, B, C;    while( scanf("%I64d%I64d%I64d",&A,&C,&B ) !=EOF )    {        if ( !A && !B && !C ) break;        memset(flag,0,sizeof(flag));        lint tmp = BabyStep_GiantStep ( A, B, C );        if ( tmp == -1 )puts("No Solution");        else printf("%I64d\n",tmp);    }    return 0;}