poj 3243 Clever Y

Clever Y

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 5744		Accepted: 1376

Description

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

X^Y 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


这题数据不强。第二种类型，C是任意的非负整数。
这个时候要用到的，不断地缩小，直到互质。代码...可以当一个模板用。
详细的报告：http://blog.csdn.net/tsaid/article/details/7354716

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
using namespace std;
const int MAX=499991;
typedef __int64 LL;
LL A,B,C;
bool Hash[MAX];
LL val[MAX];
LL idx[MAX];

LL gcd(LL a,LL b)
{
    if(b==0)
    return a;
    return gcd(b,a%b);
}

void Ex_gcd(LL a,LL b,LL &x,LL &y)
{
    if(b==0)
    {
        x=1;
        y=0;
        return ;
    }
    Ex_gcd(b,a%b,x,y);
    LL hxl=x-a/b*y;
    x=y;
    y=hxl;
    return ;
}

void Insert(LL id,LL num)//哈希建立
{
    LL k=num%MAX;
    while(Hash[k] && val[k]!=num)
    {
        k++; if(k==MAX) k=k-MAX;
    }
    if(!Hash[k])
    {
        Hash[k]=true;
        val[k]=num;
        idx[k]=id;
    }
    return;
}

LL found(LL num)//哈希查询
{
    LL k=num%MAX;
    while(Hash[k] && val[k]!=num)
    {
        k++;
        if(k==MAX) k=k-MAX;
    }
    if(Hash[k])
    {
        return idx[k];
    }
    return -1;
}

LL baby_step(LL a,LL b,LL c)
{
    LL temp=1;
    for(LL i=0;i<=100;i++)
    {
        if(temp==b%c) return i;
        temp=(temp*a)%c;
    }
    LL tmp,D=1,count=0;
    memset(Hash,false,sizeof(Hash));
    memset(val,-1,sizeof(val));
    memset(idx,-1,sizeof(idx));

    while( (tmp=gcd(a,c))!=1 )
    {
        if(b%tmp) return -1;
        c=c/tmp;
        b=b/tmp;
        D=D*a/tmp%c;
        count++;
    }
    LL cur=1;
    LL M=ceil(sqrt(c*1.0));
    for(LL i=1;i<=M;i++)
    {
        cur=cur*a%c;
        Insert(i,cur);
    }
    LL x,y;
    for(LL i=0;i<M;i++)
    {
        Ex_gcd(D,c,x,y);
        x=x*b%c;
        x=(x%c+c)%c;
        LL k=found(x);
        if(k!=-1)
        {
            return i*M+k+count;
        }
        D=D*cur%c;
    }
    return -1;

}

int main()
{
    while(scanf("%I64d%I64d%I64d",&A,&C,&B)>0)
    {
        if(A==0&&B==0&&C==0)break;
        LL cur=baby_step(A,B,C);
        if(cur==-1)
        {
            printf("No Solution
");
        }
        else printf("%I64d
",cur);
    }
    return 0;
}