Root（hdu5777+扩展欧几里得+原根）2015 Multi-University Training Contest 7

Root

Time Limit: 30000/15000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 34 Accepted Submission(s): 6


[align=left]Problem Description[/align]
Given a number sum(1≤sum≤100000000),we have m queries which contains a pair (xi,yi) and would like to know the smallest nonnegative integer kisatisfying xkii=yi mod p when the prime number p (sum mod p=0)(ps:00=1)

[align=left]Input[/align]
The first line contains a number T, indicating the number of test cases.

For each case, each case contains two integers sum,m(1≤sum≤100000000,1≤m≤100000) in the first line.

The next m lines will contains two intgeers xi,yi(0≤xi,yi≤1000000000)

[align=left]Output[/align]
For each test case,output "Case #X:" and m lines.(X is the case number)

Each line cotain a integer which is the smallest integer for (xi,yi) ,if we can't find such a integer just output "-1" without quote.

[align=left]Sample Input[/align]

1

175 2

2 1
2 3

[align=left]Sample Output[/align]

Case #1:
0
3

Hint

175 =5^2∗7

2^0 mod 5 = 1

2^3 mod 7 = 1

So the answer to (2,1) is 0

[align=left]Source[/align]
2015 Multi-University Training Contest 7

比较经典一道扩展欧几里得

现在，我们首先来解决下原根的问题：简单的解释可以参考：>>原根<<

资源下载：http://download.csdn.net/detail/u010579068/8993383

不急看懂的，可以先去切道 定义题 链接：1135 原根

解题 /article/6554554.html

转载请注明出处：寻找&星空の孩子

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5377

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<map>
using namespace std;
typedef long long ll;
const int maxn=1e6+1;
const int maxv=1e5+1;
bool isnp[maxv];
int prime[maxv],pnum;//素数数组,素数个数
int cas;
void get_prime()//素数打表
{
pnum=0;
int i,j;
memset(isnp,0,sizeof(isnp));
isnp[0]=isnp[1]=true;
for(i=2; i<maxv; i++)
{
if(!isnp[i])prime[pnum++]=i;
for(j=0; j<pnum&&prime[j]*i<maxv; j++)
{
isnp[i*prime[j]]=true;
if(i%prime[j]==0)break;
}
}
}
ll qukpow(ll k,ll base,ll p)
{
ll res=1;
for(; k; k>>=1)
{
if(k&1)res=(res*base)%p;
base=(base*base)%p;
}
return res;
}
ll ppow(ll a,ll b,ll mod)
{
ll c=1;
while(b)
{
if(b&1) c=c*a%mod;
b>>=1;
a=a*a%mod;
}
return c;
}
ll fpr[maxv];

ll find_primitive_root(ll p)//求p的原根    g^(p-1) = 1 (mod p); 求g
{
ll cnt=0,num=p-1,res;
int i;
if(p==2)return 1;
for(i=0; i<pnum && prime[i]*prime[i]<=num && num>1 ; i++)
{
if(num%prime[i]==0)//
{
fpr[cnt++]=prime[i];
while(num%prime[i]==0)num/=prime[i];
}
}
if(num>1)fpr[cnt++]=num;//fpr[]存的是p-1的因子
for(res=2; res<=p-1; res++)//暴力
{
for(i=0; i<cnt; i++)
if(ppow(res,p/prime[i],p)==1)break;
if(i>=cnt)return res;
}
return -1;
};

const int mod=1e6+7;

struct solve
{
struct HashTable
{
int top,head[mod];
struct Node
{
int x,y,next;
} node[mod];
void init()
{
top=0;
memset(head,0,sizeof(head));
}
void insert(ll x,ll y)
{
node[top].x=x;
node[top].y=y;
node[top].next=head[x%mod];
head[x%mod]=top++;
}
ll query(ll x)
{
for(int tx=head[x%mod]; tx; tx=node[tx].next)
if(node[tx].x==x)return node[tx].y;
return -1;
}
} mp;

ll p;
ll discretelog(ll prt,ll a) //取对数
{
ll res,am=ppow(prt,maxn-1,p),inv=ppow(a,p-2,p),x=1;
for(ll i=maxn-1;; i+=(maxn-1))
{
if((res=mp.query((x=x*am%p)*inv%p))!=-1)
{

return i-res;
}
if(i>p)break;
}
return -1;
}
void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y)//扩展欧几里得 x为最后需要的k
{
if(!b)
{
d=a;
x=1;
y=0;
}
else
{
ex_gcd(b,a%b,d,y,x);
y-=x*(a/b);
}
}

ll proot;
void init()
{
mp.init();
ll tmp,x,y,d;
int i;
proot=find_primitive_root(p);//找到素数p的原根
for(i=0,tmp=1; i<maxn-1; i++,tmp=tmp*proot%p)
mp.insert(tmp%p,i*1ll);
}
ll query(ll x,ll y)
{
ll d;
x%=p;
y%=p;

if(y==1)return 0;
else if(x==0)
{
if(y==0)return 1;
else return -1;
}
else if(y==0)return -1;
else
{
ll s=discretelog(proot,x);

ll t=discretelog(proot,y);

ex_gcd(s,p-1,d,x,y);
if(t%d)return -1;
else
{
ll dx=(p-1)/d;
x=(x%dx+dx)%dx;
x*=(t/d);
x=(x%dx+dx)%dx;
return x;
}
}
}
} sol[32];
int main()
{
int i,j,q,con,T;
ll sum,x,y;
scanf("%d",&T);
get_prime();
cas=1;
while(cas<=T)
{
con=0;
scanf("%I64d %d",&sum,&q);

for(i=0; i<pnum&&prime[i]*prime[i]<=sum&&sum!=1; i++)
{
if(sum%prime[i]==0)//素数存起来
{
sol[con].p=prime[i];
sol[con].init();
con++;
while(sum%prime[i]==0)sum/=prime[i];
}
}
if(sum>1)
{
sol[con].p=sum;
sol[con].init();
con++;
}

printf("Case #%d:\n",cas++);

for(i=0; i<q; i++)
{
scanf("%lld %lld",&x,&y);

ll res=1e18,tmp;
for(j=0; j<con; j++)
{

tmp=sol[j].query(x,y);
if(tmp!=-1)res=min(res,tmp);
}
if(res==1e18)res=-1;
printf("%I64d\n",res);
}
}
return 0;
}