HDU 2669 Romantic 扩展欧几里得裸题 有不理解的地方

RomanticTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4304    Accepted Submission(s): 1808Problem DescriptionThe Sky is Sprite.The Birds is Fly in the Sky.The Wind is Wonderful.Blew Throw the TreesTrees are Shaking, Leaves are Falling.Lovers Walk passing, and so are You. ................................Write in English class by yifenfeiGirls 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.InputThe input contains multiple test cases.Each case two nonnegative integer a,b (0<a, b<=2^31)Outputoutput 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 Input77 5110 4434 79Sample Output2 -3sorry7 -3AuthoryifenfeiSourceHDU女生专场公开赛——谁说女子不如男Recommendlcy   |   We have carefully selected several similar problems for you:  2668 2674 2671 2670 2672 

来源： http://acm.hdu.edu.cn/showproblem.php?pid=2669

//!  设次数为ans  ：     (ans*m+x)- (ans*n+y) = L*times//!  --> ans*(m-n) + (-L)*times = y - x#include <iostream>#include <cstdio>#include <cmath>typedef long long LL;using namespace std;void exgcd(LL a,LL b,LL &x,LL &y,LL &d){    if(!b) { d=a; x=1; y=0;}    else {exgcd(b,a%b,y,x,d);y-=x*(a/b);}}int Cal(LL a,LL b,LL c)    -->可以得到ax+by = c的最小正整数x  而y = (c-x*a)/b;{//!获得最小的x解    LL x,y,gcd;exgcd(a,b,x,y,gcd);    if(c%gcd) return -1;    x*=c/gcd;b/=gcd;    if(b<0) b=-b;    LL ans=x%b;    return ans<=0?ans+b:ans;}int main(void){   // freopen("D:\\test.txt","r",stdin);    LL a,b;    while(cin>>a>>b)    {        LL ans = Cal(a,b,1);  //为什么要化为1元求    -->ax+by=c  此处c=1        if(ans==-1) cout<<"sorry"<<endl;         else printf("%I64d %I64d\n",ans,(1-ans*a)/b);    }}