hdu 1576

设A=B*k1=k2*99973+n
则B*k1-k2*9973=n
因为gcd(B,9973)=1;
用扩展欧几里得算法可以求出x1,x2使
x1*b+x2*9973=1;
n*x1*b+n*x2*9973=n;
求(n*x1)%9973;

因为求出的x1可能有负数,(n*k%9973+9973)%9973;

#include<stdio.h>#include<math.h>#define inf 9973int q,x,y;void gcd(int a,int b){if(b==0){q=a;x=1;y=0;return;}else {gcd(b,a%b);int temp=x;x=y;y=temp-a/b*x;}}int main(){int i,j,b,n,t,k1;scanf("%d",&t);while(t--){scanf("%d%d",&n,&b);if(inf<b){gcd(b,inf);k1=x;}else{   gcd(inf,b);   k1=y;}//printf("%d=(%d)*(%d)+(%d)*(%d)\n",q,x+y-k1,inf,k1,b);k1=(n*k1)%inf;k1=(k1+inf)%inf;//求出的k1有负数，刚开始求绝对值发现不对，跟答案相加正好等于9973printf("%d\n",k1);}return 0;}