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 < 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 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
", ans);
}

return 0;

}