Time Limit: 2000/1000 MS (Java/Others)
Ignatius is poor at math, he falls across a puzzle problem, so he has no choice but to appeal to Eddy. This problem describes that:
f
(
x
)
=
5
x
13
+
13
x
5
+
k
a
x
f(x)=5x^{13}+13x^5+kax
f ( x ) = 5 x 1 3 + 1 3 x 5 + k a x
k
(
k
<
10000
)
k(k<10000)
k ( k < 1 0 0 0 0 )
a
a
a
x
x
x
65
∣
f
(
x
)
65\mid f(x)
6 5 ∣ f ( x )
a
a
a "no".
The input contains several test cases. Each test case consists of a nonegative integer
k
k
k
The output contains a string "no",if you can’t find
a
a
a
a
a
a
11
100
9999
22
no
43
#include <bits/stdc++.h>
using namespace std;
int a, k, t;
void exgcd ( int a, int b, int & x, int & y) {
if ( ! b) x= 1 , y= 0 ;
else exgcd ( b, a% b, y, x) , y- = a/ b* x;
}
int main ( ) {
while ( ~ scanf ( "%d" , & a) ) {
if ( ! ( a% 5 ) || ! ( a% 13 ) )
printf ( "no\n" ) ;
else {
exgcd ( a, 65 , k, t) ;
k= ( k* ( - 18 ) % 65 + 65 ) % 65 ;
printf ( "%d\n" , k) ;
}
}
return 0 ;
}
看了Discuss里面的解答，似乎都不够严谨，笔者试着写一下自己的过程。
65
∣
5
x
13
+
13
x
5
+
k
a
x
,
∀
x
∈
Z
,
65 \mid 5x^{13}+13x^5+kax,\forall x\in \mathbb{Z},
6 5 ∣ 5 x 1 3 + 1 3 x 5 + k a x , ∀ x ∈ Z ,
{
5
∣
13
x
5
+
k
a
x
13
∣
5
x
13
+
k
a
x
,
∀
x
∈
Z
.
\left\{\begin{aligned}5 \mid 13x^5+kax \\13 \mid 5x^{13 }+kax\end{aligned}\right.,\forall x\in \mathbb{Z}.
{ 5 ∣ 1 3 x 5 + k a x 1 3 ∣ 5 x 1 3 + k a x , ∀ x ∈ Z .
5
∣
x
5\mid x
5 ∣ x
13
∣
x
13\mid x
1 3 ∣ x
5
∤
x
5\nmid x
5 ∤ x
13
∤
x
13 \nmid x
1 3 ∤ x
{
5
∣
13
(
x
4
−
1
)
+
13
+
k
a
13
∣
5
(
x
12
−
1
)
+
5
+
k
a
.
\left\{\begin{aligned}&5 \mid 13(x^4-1)+13+ka \\&13 \mid 5(x^{12 }-1)+5+ka\end{aligned}\right..
{ 5 ∣ 1 3 ( x 4 − 1 ) + 1 3 + k a 1 3 ∣ 5 ( x 1 2 − 1 ) + 5 + k a .
{
5
∣
x
4
−
1
,
5
∤
x
13
∣
x
12
−
1
,
13
∤
x
\left\{\begin{aligned}&5 \mid x^4-1,5\nmid x\\&13 \mid x^{12}-1,13\nmid x\end{aligned}\right.
{ 5 ∣ x 4 − 1 , 5 ∤ x 1 3 ∣ x 1 2 − 1 , 1 3 ∤ x
{
5
∣
13
+
k
a
13
∣
5
+
k
a
,
\left\{\begin{aligned}5 \mid 13+ka \\13 \mid 5+ka\end{aligned}\right.,
{ 5 ∣ 1 3 + k a 1 3 ∣ 5 + k a ,
k
a
=
5
x
−
13
=
13
y
−
5
,
x
,
y
∈
Z
ka=5x-13=13y-5,x,y \in \mathbb{Z}
k a = 5 x − 1 3 = 1 3 y − 5 , x , y ∈ Z
{
x
=
−
1
−
13
t
y
=
−
1
−
5
t
k
a
=
−
18
−
65
t
,
t
∈
Z
,
\left\{\begin{aligned}&x=-1-13t \\&y=-1-5t\\&ka=-18-65t\end{aligned}\right.,t \in \mathbb{Z},
⎩ ⎪ ⎨ ⎪ ⎧ x = − 1 − 1 3 t y = − 1 − 5 t k a = − 1 8 − 6 5 t , t ∈ Z ,
a
k
+
65
t
=
−
18
,
k
,
t
∈
Z
ak+65t=-18,k,t\in \mathbb{Z}
a k + 6 5 t = − 1 8 , k , t ∈ Z
k
k
k
gcd
(
a
,
65
)
∣
−
18
\text{gcd}(a,65)|-18
gcd ( a , 6 5 ) ∣ − 1 8
gcd
(
65
,
−
18
)
=
1
\text{gcd}(65,-18)=1
gcd ( 6 5 , − 1 8 ) = 1
a
a
a
gcd
(
a
,
65
)
=
1
\text{gcd}(a,65)=1
gcd ( a , 6 5 ) = 1
a
k
+
65
t
=
1
,
k
,
t
∈
Z
ak+65t=1,k,t\in \mathbb{Z}
a k + 6 5 t = 1 , k , t ∈ Z
−
18
-18
− 1 8
k
k
k
b
y
0
+
(
a
%
b
)
x
0
=
gcd
(
b
,
a
%
b
)
⇔
a
x
0
+
b
(
y
0
−
⌊
a
b
⌋
x
0
)
=
gcd
(
a
,
b
)
.
by_0+(a\%b)x_0=\gcd(b,a\%b)\\ \Leftrightarrow ax_0+b(y_0-\lfloor \frac{a}{b} \rfloor x_0)=\gcd(a,b).
b y 0 + ( a % b ) x 0 = g cd( b , a % b ) ⇔ a x 0 + b ( y 0 − ⌊ b a ⌋ x 0 ) = g cd( a , b ) .
1
1
1
65
65
6 5
k
k
k
65
∣
a
k
+
18.
65 \mid ak+18.
6 5 ∣ a k + 1 8 .