大小步(同餘方程的解)
阿新 • • 發佈:2019-02-03
大小步BigStepGiantStep演算法求 A^x = B( mod P ) (注:P為質數)中x的解
//來自kuangbin的ACM模板
#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為素數,a,b < n
// 求解上式 0<=x < n的解
#define 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;
long long x = 1, p = 1 ;
for(int i = 0; i < m; ++i, p = p*a%n)insert(p*b%n,i);
for(long long i = m; ;i += m)
{
if( (j = find(x = x*p%n)) != -1 )return i-j;
if(i > n)break;
}
return -1;
}
int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);
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;
}
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