poj3243

maksyuki 发表于 oj 分类，标签: 数论-离散对数

15 9月 2015

Clever Y

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

No Solution

Source

POJ Monthly--2007.07.08, Guo, Huayang

题目类型：形如 A ^ X = B (mod C)的高次同余方程

算法分析：直接按照A ^ X = B (mod C)的格式求解即可

#include <set>
#include <bitset>
#include <list>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#include <ios>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <algorithm>
#include <utility>
#include <complex>
#include <numeric>
#include <functional>
#include <cmath>
#include <ctime>
#include <climits>
#include <cstdarg>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <cassert>

#define lson rt << 1, l, m
#define rson rt << 1 | 1, m + 1, r

using namespace std;

const int INF = 0x7FFFFFFF;
const double EPS = 1e-10;
const double PI = 2 * acos (0.0);
const int MOD = 7;
const int maxn = 110000 + 66;

struct HashNode
{
    long long data, id, next;
};

HashNode Hash[maxn<<1];
bool flag[maxn<<1];
long long top;

void Insert (long long a, long long b)
{
    long long 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;
}

long long Find (long long b)
{
    long long 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;
}

long long gcd (long long a, long long b)
{
    return b ? gcd (b, a % b) : a;
}

long long gcd_ex (long long a, long long b, long long &x, long long &y)
{
    if (b == 0) { x = 1; y = 0; return a; }
    long long d = gcd_ex (b, a % b, y, x);
    y = y - a / b * x;
    return d;
}


long long q_mul_mod (long long a, long long b, long long p)
{
    long long ans = 1;
    while (b)
    {
        if (b & 1)
        {
            ans = ((ans % p) * (a % p)) % p;
            b--;
        }

        b = b >> 1;
        a = ((a % p) * (a % p)) % p;
    }

return ans;
}

long long BabyStep_GiantStep (long long A, long long B, long long C)
{
    top = maxn;  B %= C;
    long long tmp = 1, i;

    for (i = 0; i <= 100; tmp = tmp * A % C, i++)
        if (tmp == B % C)
            return i;

    long long 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++;
    }

    long long M = (long long) ceil (sqrt (C + 0.0));

    for (tmp = 1, i = 0; i <= M; tmp = tmp * A % C, i++)
        Insert (i, tmp);

    long long x, y, K = q_mul_mod (A, M, C);
    for (i = 0; i <= M; i++)
    {
        gcd_ex (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()
{
//    ifstream cin ("aaa.txt");

    long long A, B, C;
    while(cin >> A >> C >> B)
    {
        if (A == 0 && B == 0 && C == 0)
            break;

        memset (flag, 0, sizeof(flag));
        long long ans = BabyStep_GiantStep (A, B, C);

        if (ans == -1)
            cout << "No Solution" << endl;

        else
            cout << ans << endl;
    }

return 0;
}

100

101

102

103

104

105

106

107

108

109

110

111

112

113

114

115

116

117

118

119

120

121

122

123

124

125

126

127

128

129

130

131

132

133

134

135

136

137

138

139

140

141

142

143

144

145

146

147

148

149

150

151

152

153

154

155

156

157

158

159

160

161

162

163

164

165

166

167

168

169

170

171

172

173

174

175

176

177

178

179

180

181

182

183

184

185

186

187

188

#include <set>

#include <bitset>

#include <list>

#include <map>

#include <stack>

#include <queue>

#include <deque>

#include <string>

#include <vector>

#include <ios>

#include <iostream>

#include <fstream>

#include <sstream>

#include <iomanip>

#include <algorithm>

#include <utility>

#include <complex>

#include <numeric>

#include <functional>

#include <cmath>

#include <ctime>

#include <climits>

#include <cstdarg>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <cctype>

#include <cassert>

#define lson rt << 1, l, m

#define rson rt << 1 | 1, m + 1, r

using namespace std;

const int INF = 0x7FFFFFFF;

const double EPS = 1e-10;

const double PI = 2 * acos (0.0);

const int MOD = 7;

const int maxn = 110000 + 66;

struct HashNode

{

long long data, id, next;

};

HashNode Hash[maxn<<1];

bool flag[maxn<<1];

long long top;

void Insert (long long a, long long b)

{

long long 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;

}

long long Find (long long b)

{

long long 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;

}

long long gcd (long long a, long long b)

{

return b ? gcd (b, a % b) : a;

}

long long gcd_ex (long long a, long long b, long long &x, long long &y)

{

if (b == 0) { x = 1; y = 0; return a; }

long long d = gcd_ex (b, a % b, y, x);

y = y - a / b * x;

return d;

}

long long q_mul_mod (long long a, long long b, long long p)

{

long long ans = 1;

while (b)

{

if (b & 1)

{

ans = ((ans % p) * (a % p)) % p;

b--;

}

b = b >> 1;

a = ((a % p) * (a % p)) % p;

}

return ans;

}

long long BabyStep_GiantStep (long long A, long long B, long long C)

{

top = maxn; B %= C;

long long tmp = 1, i;

for (i = 0; i <= 100; tmp = tmp * A % C, i++)

if (tmp == B % C)

return i;

long long 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++;

}

long long M = (long long) ceil (sqrt (C + 0.0));

for (tmp = 1, i = 0; i <= M; tmp = tmp * A % C, i++)

Insert (i, tmp);

long long x, y, K = q_mul_mod (A, M, C);

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

{

gcd_ex (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()

{

// ifstream cin ("aaa.txt");

long long A, B, C;

while(cin >> A >> C >> B)

{

if (A == 0 && B == 0 && C == 0)

break;

memset (flag, 0, sizeof(flag));

long long ans = BabyStep_GiantStep (A, B, C);

if (ans == -1)

cout << "No Solution" << endl;

else

cout << ans << endl;

}

return 0;

}

poj3233

poj3250

超客的水墨烟雨

poj3243

分类目录

标签云

书签

近期评论

2024年12月
一	二	三	四	五	六	日
« 6月
						1
2	3	4	5	6	7	8
9	10	11	12	13	14	15
16	17	18	19	20	21	22
23	24	25	26	27	28	29
30	31