欧几里德
学习自http://blog.csdn.net/lulipeng_cpp/article/details/7612490
/*
题解:
1.
n = A%9973;
n = A-a/9973*9973;
设:B*x=A;
n = B*x - a/9973*9973,n = B*x-9973y;
2.
问题就转化为求X.
gcd(B,9973) = B*x1+9973y1 = 1;
同乘N,B*(n*x1) - 9973*(-y1*n)=n;
即 x = n*x1;
3.为了防止x为负数,k = (x%9973 + 9973)%9973;
*/
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<stdio.h> #include<string.h> int exgcd(int a,int b,int &x,int &y) { if(b == 0) { x = 1; y = 0; return a; } int r = exgcd(b,a%b,x,y); int t = x; x = y; y = t-(a/b)*y; return r; } int main() { int n,B,x,y,t; int i,j,k; while(~scanf("%d",&t)) { while(t --) { scanf("%d %d",&n,&B); k = exgcd(B,9973,x,y); x *= n; k = (x%9973 + 9973)%9973; printf("%d\n",k); } } return 0; }