hdu 2669(扩展欧几里得)

Romantic

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4400 Accepted Submission(s): 1852


[align=left]Problem Description[/align]
The Sky is Sprite.
The Birds is Fly in the Sky.
The Wind is Wonderful.
Blew Throw the Trees
Trees are Shaking, Leaves are Falling.
Lovers Walk passing, and so are You.
................................Write in English class by yifenfei



Girls are clever and bright. In HDU every girl like math. Every girl like to solve math problem!
Now
tell you two nonnegative integer a and b. Find the nonnegative integer X
and integer Y to satisfy X*a + Y*b = 1. If no such answer print "sorry"
instead.

[align=left]Input[/align]
The input contains multiple test cases.
Each case two nonnegative integer a,b (0<a, b<=2^31)

[align=left]Output[/align]
output
nonnegative integer X and integer Y, if there are more answers than the
X smaller one will be choosed. If no answer put "sorry" instead.

[align=left]Sample Input[/align]

77 51
10 44
34 79

[align=left]Sample Output[/align]

2 -3
sorry
7 -3

题解:由ax+by = gcd(a,b) 当 a,b互素是才会有解。然后X要是尽量小的正数，假设我们得到的是 X0 ，在有解的情况下方程ax+by=c的通解为 {X0*(c/d)+k*b/gcd(a,b)} (k = ..-2,-1,0,1,2..)
所以我们可以得到在最小的X为 (X0%b+b)%b.代入得Y

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
typedef long long LL;

LL extend_gcd(LL a,LL b,LL &x,LL &y){
if(!b){
x=1,y = 0;
return a;
}else{
LL x1,y1;
LL d = extend_gcd(b,a%b,x1,y1);
x = y1;
y = x1 - a/b*y1;
return d;
}
}
int main()
{
LL a,b,x,y;
while(~scanf("%lld%lld",&a,&b)){
LL d = extend_gcd(a,b,x,y);
if(d!=1){
printf("sorry\n");
}else{
x = (x%b+b)%b;
y = (1-a*x)/b;
printf("%lld %lld\n",x,y);
}
}
return 0;
}