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  ≤ 10 ⁹).  
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

  1 #include<iostream>

  2 #include<cstdio>

  3 #include<cstdlib>

  4 #include<cstring>

  5 #include<cmath>

  6 using namespace std;

  7 const int MAX=499991;

  8 typedef __int64 LL;

  9 LL A,B,C;

 10 bool Hash[MAX];

 11 LL val[MAX];

 12 LL idx[MAX];

 13 

 14 LL gcd(LL a,LL b)

 15 {

 16     if(b==0)

 17     return a;

 18     return gcd(b,a%b);

 19 }

 20 

 21 void Ex_gcd(LL a,LL b,LL &x,LL &y)

 22 {

 23     if(b==0)

 24     {

 25         x=1;

 26         y=0;

 27         return ;

 28     }

 29     Ex_gcd(b,a%b,x,y);

 30     LL hxl=x-a/b*y;

 31     x=y;

 32     y=hxl;

 33     return ;

 34 }

 35 

 36 void Insert(LL id,LL num)//哈希建立

 37 {

 38     LL k=num%MAX;

 39     while(Hash[k] && val[k]!=num)

 40     {

 41         k++; if(k==MAX) k=k-MAX;

 42     }

 43     if(!Hash[k])

 44     {

 45         Hash[k]=true;

 46         val[k]=num;

 47         idx[k]=id;

 48     }

 49     return;

 50 }

 51 

 52 LL found(LL num)//哈希查询

 53 {

 54     LL k=num%MAX;

 55     while(Hash[k] && val[k]!=num)

 56     {

 57         k++;

 58         if(k==MAX) k=k-MAX;

 59     }

 60     if(Hash[k])

 61     {

 62         return idx[k];

 63     }

 64     return -1;

 65 }

 66 

 67 LL baby_step(LL a,LL b,LL c)

 68 {

 69     LL temp=1;

 70     for(LL i=0;i<=100;i++)

 71     {

 72         if(temp==b%c) return i;

 73         temp=(temp*a)%c;

 74     }

 75     LL tmp,D=1,count=0;

 76     memset(Hash,false,sizeof(Hash));

 77     memset(val,-1,sizeof(val));

 78     memset(idx,-1,sizeof(idx));

 79 

 80     while( (tmp=gcd(a,c))!=1 )

 81     {

 82         if(b%tmp) return -1;

 83         c=c/tmp;

 84         b=b/tmp;

 85         D=D*a/tmp%c;

 86         count++;

 87     }

 88     LL cur=1;

 89     LL M=ceil(sqrt(c*1.0));

 90     for(LL i=1;i<=M;i++)

 91     {

 92         cur=cur*a%c;

 93         Insert(i,cur);

 94     }

 95     LL x,y;

 96     for(LL i=0;i<M;i++)

 97     {

 98         Ex_gcd(D,c,x,y);

 99         x=x*b%c;

100         x=(x%c+c)%c;

101         LL k=found(x);

102         if(k!=-1)

103         {

104             return i*M+k+count;

105         }

106         D=D*cur%c;

107     }

108     return -1;

109 

110 }

111 

112 int main()

113 {

114     while(scanf("%I64d%I64d%I64d",&A,&C,&B)>0)

115     {

116         if(A==0&&B==0&&C==0)break;

117         LL cur=baby_step(A,B,C);

118         if(cur==-1)

119         {

120             printf("No Solution\n");

121         }

122         else printf("%I64d\n",cur);

123     }

124     return 0;

125 }