2019牛客暑期多校训练营（第九场）B Quadratic equation 二次剩余

题目链接

样例输入：

10
4 4
5 6
10 10
10 25
20000 100000000
0 5
3 6
220 284
0 1
1000000000 1000000000

输出：

2 2
2 3
-1 -1
5 5
10000 10000
474848249 525151758
352077071 647922939
448762649 551237578
-1 -1
366417496 633582504

解方程组，在这里我是先求y，再求x；
怎么套二次剩余模板呢；
用判断一元二次方程的：b^2-4ac来套模板，看得到的答案是不是大于等于0的，如果是就有解，否则没有

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<string>
#include<cmath>
#include<cstring>
#include<set>
#include<queue>
#include<stack>
#include<ctime>
#include<map>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define random(a,b) (rand()%(b-a+1)+a)
typedef long long ll;
using namespace std;
const int N=1e5+10;
const int INF=0x3f3f3f3f;

const int p=1e9+7;
struct hh{
	ll x,y;
	hh(){};
	hh(ll _x,ll _y){
		x=_x;y=_y;
	}
};
ll w;
hh mul(hh a,hh b,ll p){
	hh ans;
	ans.x=(a.x*b.x%p+a.y*b.y%p*w%p)%p;
	ans.y=(a.x*b.y%p+a.y*b.x%p)%p;
	return ans;
}
hh quick1(hh a,ll b,ll p){
	hh ans=hh(1,0);
	while(b){
		if(b&1) ans=mul(ans,a,p);
		a=mul(a,a,p);
		b>>=1;
	}
	return ans;
}
ll quick2(ll a,ll b,ll p){
	ll ans=1;
	while(b){
		if(b&1) ans=(ans*a)%p;
		b>>=1;
		a=(a*a)%p;
	}
	return ans;
}
ll solve(ll a,ll p){//求解 x^2=a(mod p) 的x的值
	a%=p;//注意这句话
	if(a==0) return 0;//注意这句话
	if(p==2) return a;
	if(quick2(a,(p-1)/2,p)==p-1) return -1;
	ll b,t;
	while(1){
		b=rand()%p;
		t=b*b-a;
		w=(t%p+p)%p;
		if(quick2(w,(p-1)/2,p)==p-1) break;
	}
	hh ans=hh(b,1);
	ans=quick1(ans,(p+1)/2,p);
	return ans.x;
}

ll inv(ll base,ll num){
    ll ans=1;
    while(num){
        if(num&1)
            ans=(ans*base)%p;
        num>>=1;
        base=(base*base)%p;
    }
    return ans;
}
int main()
{
    #ifndef ONLINE_JUDGE
    freopen("in.txt","r",stdin);
    #endif // ONLINE_JUDGE
    int T;
    scanf("%d",&T);
    ll b,c;
    while(T--){
        scanf("%lld%lld",&b,&c);
        ll t=b*b-4ll*c;
        t=(t+p)%p;
        t= solve(t,p);
        if(t<0){
            printf("-1 -1\n");
        }else{
            ll x=((b+t)%p*inv(2,p-2))%p;
            ll y=(b-x+p)%p;
            if(x<=y)
                printf("%lld %lld\n",x,y);
            else
                printf("%lld %lld\n",y,x);
        }
    }
    return 0;
}

模板

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<string>
#include<cmath>
#include<cstring>
#include<set>
#include<queue>
#include<stack>
#include<ctime>
#include<map>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define random(a,b) (rand()%(b-a+1)+a)
typedef long long ll;
using namespace std;
const int N=1e5+10;
const int INF=0x3f3f3f3f;

const int p=1e9+7;
struct hh{
	ll x,y;
	hh(){};
	hh(ll _x,ll _y){
		x=_x;y=_y;
	}
};
ll w;
hh mul(hh a,hh b,ll p){
	hh ans;
	ans.x=(a.x*b.x%p+a.y*b.y%p*w%p)%p;
	ans.y=(a.x*b.y%p+a.y*b.x%p)%p;
	return ans;
}
hh quick1(hh a,ll b,ll p){
	hh ans=hh(1,0);
	while(b){
		if(b&1) ans=mul(ans,a,p);
		a=mul(a,a,p);
		b>>=1;
	}
	return ans;
}
ll quick2(ll a,ll b,ll p){
	ll ans=1;
	while(b){
		if(b&1) ans=(ans*a)%p;
		b>>=1;
		a=(a*a)%p;
	}
	return ans;
}
ll solve(ll a,ll p){//求解 x^2=a(mod p) 的x的值
	a%=p;//注意这句话
	if(a==0) return 0;//注意这句话
	if(p==2) return a;
	if(quick2(a,(p-1)/2,p)==p-1) return -1;
	ll b,t;
	while(1){
		b=rand()%p;
		t=b*b-a;
		w=(t%p+p)%p;
		if(quick2(w,(p-1)/2,p)==p-1) break;
	}
	hh ans=hh(b,1);
	ans=quick1(ans,(p+1)/2,p);
	return ans.x;
}

ll inv(ll base,ll num){
    ll ans=1;
    while(num){
        if(num&1)
            ans=(ans*base)%p;
        num>>=1;
        base=(base*base)%p;
    }
    return ans;
}
int main()
{
       
       ll t= solve(t,p);
            return 0;
}