HDOJ 1576 A/B

思路：设(A/B)%9973 = K, 则A/B = k + 9973x  (x未知）， 因此A = kB + 9973xB,

又A%9973 = n， 所以kB%9973 = n,  故kB = n + 9973y (y未知)

故(k/n)B +(-y/n)*9973 = gcd(B,9973) = 1

扩展欧几里得 求出k/n,  再乘以个n，记得取模，就是answer了

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#define m 9973using namespace std;typedef long long LL;void gcd(LL a,LL b,LL &x,LL &y){    if(!b) {x=1;y=0;}    else {gcd(b,a%b,y,x);y-=x*(a/b);}}int main(){    int t;    LL n,b,x,y;    scanf("%d",&t);    while(t--)    {        scanf("%lld%lld",&n,&b);        gcd(b,m,x,y);        x=(x%m+m)%m;        printf("%lld\n",(x*n)%9973);    }    return 0;}