UVALive Problem 7457 Discrete Logarithm Problem——Regionals 2015 :: Asia - Taipei

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UVALive Problem 7457 Discrete Logarithm Problem

Accept: 0 Submit: 0
Time Limit: 3.000 seconds

Problem Description

Finite groups are used in almost all modern cryptosystems. Let p be a prime. The finite multiplicative group constructed by integers modulo p is denoted by. The elements of the group are 1, 2, . . . , p−1. Let a and b be two elements in. The value of a×b is define as a·b mod p. For example, let p = 13, a = 5, and b = 7. Then 5 × 7 = 5·7 mod 13 = 35 mod 13 = 9.

You are going to write a program to solve the discrete logarithm problem in. That is, given p, a, and b, compute x satisfying

For very large p, solving discrete logarithm problem in may not be easy. However, in this problem the value of p will not be very large.

Input

The first line of the input file contains a prime p, 1 < p < 2^13 . This prime will be used in the following test cases. Each test case contains two integers a and b in a line. The last test case is followed by a line containing ‘0’.

Output

For each test case, print out the solution x to the equation in the finite group. If there are no solutions, print ‘0’.

Sample Input

Sample Output

7
21

Problem Idea

解题思路:

【题意】
求满足的x值,若无解则输出'0'

【类型】
暴力

【分析】
其实此题已经把题目简化了

因为它给了你x的范围为[1,p-1],然而p的值又很小,所以暴力枚举x的每种可能取值,判断等式是否成立即可

另外,我们提一提如果不给你x的范围应该怎么做

由费马小定理可得

这样我们就可以把x缩小至范围[0,p-2],然后暴力枚举就可以了

因此,此题暴力枚举也无需枚举到p-1,到p-2就可以了,若到p-2还没有满足的解x,那就是无解的情况了

此外要提及的一点是,输入除了单行'0'结束外,还要判断是否到文件尾(EOF),不然会超时(TLE)

【时间复杂度&&优化】
O(p)

题目链接→UVALive Problem 7457 Discrete Logarithm Problem

Source Code

/*Sherlock and Watson and Adler*/#pragma comment(linker, "/STACK:1024000000,1024000000")#include<stdio.h>#include<string.h>#include<stdlib.h>#include<queue>#include<stack>#include<math.h>#include<vector>#include<map>#include<set>#include<bitset>#include<cmath>#include<complex>#include<string>#include<algorithm>#include<iostream>#define eps 1e-9#define LL long long#define PI acos(-1.0)#define bitnum(a) __builtin_popcount(a)using namespace std;const int N = 15;const int M = 100005;const int inf = 1000000007;const int mod = 7;int main(){    int p,a,b,i,s;    scanf("%d",&p);    while(~scanf("%d",&a)&&a)//'~'不能少,否则TLE    {        s=a%p;        scanf("%d",&b);        for(i=1;i<=p-2;i++,s=a%p*s%p)            if(s==b)                break;        if(i>p-2)            puts("0");        else            printf("%d\n",i);    }    return 0;}

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