URAL 1123 Square Root（计算二次剩余）

5
4 17
3 7
2 7
14 31
10007 20011

2 15
No root
3 4
13 18
5382 14629

#include <cstdio>
#include <algorithm>
#include <iostream>

using namespace std;
/*********************************
计算二次剩余模版即x^2 = n(mod)p
*********************************/
typedef long long ll;
ll tmod;
ll mod(ll n,ll p)
{
ll temp = 1;
while(p > 0)
{
if(p & 1) temp = temp * n % tmod;
n = n * n % tmod;
p >>= 1;
}
return temp;
}
ll pows(ll n)
{
ll i;
ll temp = 1;
for(i = 0; i < n; i++)
temp *= 2;
return temp;
}

int main()
{
int n,p,cs;
ll z,q,s,c;
ll r,t,m,b,i;
ll minx;
scanf("%d",&cs);
while(cs--)
{
scanf("%d%d",&n,&p);
//p=2时特判
if(p == 2)
{
if(n % p == 1)printf("1\n");
else printf("No root\n");
continue;
}
//如果无解
tmod = p;
if(mod(n,(p - 1) / 2) != 1)
{
printf("No root\n");
continue;
}
q = p - 1;
s = 0;
while(q % 2 == 0)
{
q /= 2;
s ++;
}
if(s == 1)
{
r = mod(n,(p + 1) / 4);
}
else
{
while(1)
{
z = 1 + rand() % (p - 1);
if(mod(z,(p - 1) / 2) == (p - 1)) break;
}
c = mod(z,q);
r = mod(n,(1 + q) / 2);
t = mod(n,q);
m = s;
while(1)
{
if(t % p == 1) break;
for(i = 1; i < m; i++)
{
if(mod(t,pows(i)) == 1) break;
}
b = mod(c, pows(m - i - 1));
r = r * b % p;
t = t * b * b % p;
c = b * b % p;
m = i;
}
}
r = (r % p + p) % p;
//如果只有一个
if(r == p - r)cout<<r<<endl;
//如果有两个按照升序输出
else
{
minx = min(r,p - r);
cout<<minx<<" "<<(p - minx)<<endl;
}
}
return 0;
}