你被要求设计一个计算器完成以下三项任务:
1、给定y,z,p,计算Y^Z Mod P 的值;(快速幂)
2、给定y,z,p,计算满足xy≡ Z ( mod P )的最小非负整数;(exgcd)
3、给定y,z,p,计算满足Y^x ≡ Z ( mod P)的最小非负整数。(BSGS)
/**************************************************************
Problem: 2242
User: walfy
Language: C++
Result: Accepted
Time:2244 ms
Memory:3820 kb
****************************************************************/
//#pragma comment(linker, "/stack:200000000")
//#pragma GCC optimize("Ofast,no-stack-protector")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define pi acos(-1.0)
#define ll long long
#define vi vector<int>
#define mod 1000000007
#define C 0.5772156649
#define ls l,m,rt<<1
#define rs m+1,r,rt<<1|1
#define pil pair<int,ll>
#define pli pair<ll,int>
#define pii pair<int,int>
#define cd complex<double>
#define ull unsigned long long
#define base 1000000000000000000
#define fio ios::sync_with_stdio(false);cin.tie(0)
using namespace std;
const double g=10.0,eps=1e-12;
const int N=100000+10,maxn=1000+10,inf=0x3f3f3f3f,INF=0x3f3f3f3f3f3f3f3f;
inline ll quick(ll a,ll b,ll p)
{
ll ans=1;
while(b)
{
if(b&1)ans=ans*a%p;
a=a*a%p;b>>=1;
}
return ans;
}
inline ll exgcd(ll a,ll b,ll &x,ll &y)
{
if(!b){x=1,y=0;return a;}
ll ans=exgcd(b,a%b,x,y);
ll t=x;x=y;y=t-a/b*y;
return ans;
}
inline ll eg(ll y,ll z,ll p)
{
ll x,k;
exgcd(y,p,x,k);
x=x*z/__gcd(y,p);
ll b=p/__gcd(y,p);
x=(x%b+b)%b;
return x;
}
map<ll,ll>ma;
int main()
{
int n,op;
while(~scanf("%d%d",&n,&op))
{
ll y,z,p;
for(int i=0;i<n;i++)
{
scanf("%lld%lld%lld",&y,&z,&p);
if(op==1)printf("%lld\n",quick(y,z,p));
else if(op==2)
{
if(z%__gcd(y,p)==0)printf("%lld\n",eg(y,z,p));
else puts("Orz, I cannot find x!");
}
else
{
ll a=y,b=z;
if(a%p==0)
{
puts("Orz, I cannot find x!");
continue;
}
ma.clear();
ll m=ceil(sqrt(p));
ll now=b%p;
ma[now]=0;
for(int i=1;i<=m;i++)
{
now=(now*a)%p;
ma[now]=i;
}
ll ans=-1,t=quick(a,m,p);now=1;
for(int i=1;i<=m;i++)
{
now=now*t%p;
if(ma[now])
{
ans=i*m-ma[now];
break;
}
}
if(ans!=-1)printf("%lld\n",(ans%p+p)%p);
else puts("Orz, I cannot find x!");
}
}
}
return 0;
}
/***********************
***********************/
原文地址:https://www.cnblogs.com/acjiumeng/p/9276012.html
时间: 2024-10-24 21:55:27