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Quadratic equation（二次剩余）2019牛客多校第九场
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鏈接：https://ac.nowcoder.com/acm/contest/889/B
題目描述
Let p = 1000000007.
In the following t lines, each line contains two integers b and c (0 <= b, c < p).
If there is no solution, please output -1, -1
#include <bits/stdc++.h>
typedef long long ll
;
using namespace std
;
ll
pow_mod ( ll a
, ll n
, ll p
) { ll res
= 1 ; while ( n
) { if ( n
& 1 ) res
= res
* a
% p
; n
>>= 1 ; a
= a
* a
% p
; } return res
;
}
ll
legendre ( ll a
, ll p
) { return pow_mod ( a
, ( p
- 1 ) >> 1 , p
) ;
}
ll
mod ( ll a
, ll p
) { a
% = p
; if ( a
< 0 ) a
+ = p
; return a
;
}
struct node
{ static ll p
, omega
; ll a
, b
; node ( ll a
, ll b
) : a ( a
% p
) , b ( b
% p
) { }
} ;
node
operator * ( const node
& p
, const node
& q
) { int m
= node
:: p
; return node ( p
. a
* q
. a
+ p
. b
* q
. b
% m
* node
:: omega
, p
. a
* q
. b
+ q
. a
* p
. b
) ;
}
node
pow_mod ( node a
, ll n
) { node
result ( 1 , 0 ) ; while ( n
> 0 ) { if ( ( n
& 1 ) == 1 ) { result
= result
* a
; } a
= a
* a
; n
>>= 1 ; } return result
;
}
ll node
:: p
, node
:: omega
;
ll
modsqr ( ll a
, ll p
) { if ( p
== 2 ) return 1 ; if ( legendre ( a
, p
) + 1 == p
) return - 1 ; if ( ( ( ( p
+ 1 ) >> 1 ) & 1 ) == 0 ) return pow_mod ( a
, ( p
+ 1 ) >> 2 , p
) ; ll a_0
= - 1 ; while ( true ) { a_0
= rand ( ) % p
; if ( legendre ( mod ( a_0
* a_0
- a
, p
) , p
) + 1 == p
) break ; } node
:: p
= p
; node
:: omega
= mod ( a_0
* a_0
- a
, p
) ; node ret
= pow_mod ( node ( a_0
, 1 ) , ( p
+ 1 ) >> 1 ) ; return ret
. a
;
}
const long long mood
= 1000000007 ;
int main
( )
{ ll b
, c
; ll s2
= pow_mod ( 2 , mood
- 2 , mood
) ; int T
; scanf ( "%d" , & T
) ; while ( T
-- ) { scanf ( "%lld %lld" , & b
, & c
) ; ll t
= b
* b
- 4 * c
; t
= t
% mood
; ll x_y
= modsqr ( t
, mood
) ; if ( x_y
== - 1 ) { printf ( "-1 -1\n" ) ; } else { if ( x_y
* 2 > mood
) x_y
= mood
- x_y
; ll x
= ( ( ( x_y
+ b
) % mood
) * s2
) % mood
; ll y
= ( ( b
- x
) + mood
) % mood
; if ( x
> y
) swap ( x
, y
) ; if ( ( x
+ y
) % mood
== b
&& ( x
* y
) % mood
== c
) printf ( "%lld %lld\n" , x
, y
) ; else printf ( "-1 -1\n" ) ; } } return 0 ;
}
努力加油a啊，(o )/~
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