Romantic

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 5165    Accepted Submission(s): 2163 Problem 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.   Input The input contains multiple test cases. 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   题意：给出非负整数a，b；求的满足a*x+b*y=1的非负整数x和整数y，若没有则输出sorry 题解：今天重新看了扩展欧几里德，基本是懂了，半年前就看过了，一直不懂，智商啄鸡.... x与y解的通项式：   x=x 0 +(b/gcd)*k      y=y 0 -(a/gcd)*k  (k为正整数，gcd为gcd(a,b) ) 证明见刘汝佳紫书P.313 裸题，不赘述。 代码如下： #include<cstdio> #include<cstring> #include<algorithm> using namespace std; #define LL long long void exgcd(LL a,LL b,LL &d,LL &x,LL &y) { if(b==0) { x=1; y=0; d=a; return ; } else { exgcd(b,a%b,d,y,x); y-=(a/b)*x; } } int main() { LL a,b,d,x,y; while(scanf("%lld%lld",&a,&b)!=EOF) { exgcd(a,b,d,x,y); if(d!=1) printf("sorry\n"); else{ x=x%b; if(x<=0) x+=b; printf("%lld %lld\n",x,(1-x*a)/b); } } return 0; }