UVALive 7457 Discrete Logarithm Problem （暴力枚举）

Discrete Logarithm Problem

题目链接：

http://acm.hust.edu.cn/vjudge/contest/127401#problem/D

Description

http://7xjob4.com1.z0.glb.clouddn.com/42600d1bedc3c308a68ea0453befd84e

Input

The first line of the input file contains a prime p, 1

Output

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

Sample Input

31
24 3
3 15
0

Sample Output

7
21


##题意：

求一个x使得 a^x%p = b .


##题解：

由于p的规模巨小(2^13)，而取模的结果一定在p次以内出现循环，所以直接从0~p枚举x即可.
很多人TLE了，不知道什么姿势. (可能是末尾的0没有读好)
虽然是水题，还是很高兴拿了FB.


##代码：
``` cpp
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define LL long long
#define eps 1e-8
#define maxn 1010
#define mod 100000007
#define inf 0x3f3f3f3f
#define mid(a,b) ((a+b)>>1)
#define IN freopen("in.txt","r",stdin);
using namespace std;

LL quickmod(LL a,LL b,LL m)

{

LL ans = 1;

while(b){

if(b&1){

ans = (ansa)%m;

b--;

}

b/=2;

a = aa%m;

}

return ans;

}

int main(int argc, char const *argv[])

{

//IN;

int p; cin >> p;
LL a, b;
while(scanf("%lld", &a) != EOF && a)
{
    scanf("%lld", &b);
    a %= p; //b %= p;

    int ans = 0;
    for(int i=0; i<=p+1; i++) {
        if(quickmod(a, i, p) == b) {
            ans = i;
            break;
        }
    }

    printf("%d\n", ans);
}

return 0;

}