poj 2417 Discrete Logging

转载请注明出处，谢谢http://blog.csdn.net/bigtiao097?viewmode=contents

题意：

给定A、B、C，求

Ax≡B(modC)

的最小x值
这题的C一定是质数

思路：

模板题,BSGS算法

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具体代码如下：
Result：Accepted    Memory: 1492K   Time : 32MS

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <math.h>#include <stdlib.h>#include <time.h>using namespace std;//baby_step giant_step// a^x = b (mod n) n为素数，a,b < n// 求解上式 0<=x < n的解typedef long long ll;const int MOD = 76543;int hs[MOD],head[MOD],next[MOD],id[MOD],top;void insert(int x,int y){    int k = x%MOD;    hs[top] = x, id[top] = y, next[top] = head[k], head[k] = top++;}int find(int x){    int k = x%MOD;    for(int i = head[k]; i != -1; i = next[i])        if(hs[i] == x)            return id[i];    return -1;}int BSGS(int a,int b,int n){    memset(head,-1,sizeof(head));    top = 1;    if(b == 1)return 0;    int m = sqrt(n*1.0), j;    ll x = 1, p = 1;    for(int i = 0; i < m; ++i, p = p*a%n)insert(p*b%n,i);    for(ll i = m; ;i += m)    {        if( (j = find(x = x*p%n)) != -1 )return i-j;        if(i > n)break;    }    return -1;}int main(){    int a,b,n;    while(scanf("%d%d%d",&n,&a,&b) == 3)    {        int ans = BSGS(a,b,n);        if(ans == -1)printf("no solution\n");        else printf("%d\n",ans);    }    return 0;}