Description
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.
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.
Input
The input contains multiple test cases.
Each case two nonnegative integer a,b (0<a, b<=2^31)
Each case two nonnegative integer a,b (0<a, b<=2^31)
Output
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.
Sample Input
77 51 10 44 34 79
Sample Output
2 -3 sorry 7 -3
分析：ax+by=gcd(a,b)*k，只有k为整数时，才有解，所以当前仅当gcd(a,b)=1时，这道题有解。
当求解出x<0时，利用
(x+b/gcd(a,b),y-a/gcd(a,b) 都是符合条件的解这一性质，找到第一个符合条件的x输出之。
#include<iostream> #include<stdio.h> using namespace std; long long ext_gcd(long long a,long long b,long long *x,long long *y) { if(b==0) { *x=1; *y=0; return a; } long long r = ext_gcd(b,a%b,x,y); long long t = *x; *x=*y; *y=t - a/b * *y; return r; } int main() { long long a,b,x,y; while(~scanf("%I64d%I64d",&a,&b)) { if(ext_gcd(a,b,&x,&y)==1) { while(x<0) { x+=b/1; y-=a/1; } printf("%I64d %I64d ",x,y); continue; } else printf("sorry "); } return 0; }