Hdu 1576 A/B

放假后回来第2天在网吧AC的第一道题。这是一道变形的扩展欧几里得算法题。由题知：n = A%9973 。设9973*y = n;设A/B = x,则A = B*x;  则题目可以转换为: B*x-9973*y = n;对于扩展欧几里得算法的使用还不熟练，要加强联系啦！~

CODE:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
using namespace std;

int n, b;

int ex_gcd(int a, int b, int &x, int &y)    
{
    if(!b)
    {
        x = 1; y = 0; return a;
    }
    else
    {
        int d = ex_gcd(b, a%b, x, y);
        int t = x;
        x = y;
        y = t-(a/b)*y;
        return d;
    }
} //ex_gcd


int main()
{
    int T;
    int a, b;
    int x, y, k;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d%d", &n, &b);
        int d = ex_gcd(b, 9973, x, y);
        x *= n/d;
        k = 9973/d;
        x = (x%k+k)%k;               //满足条件的最小值 
        printf("%d\n", x%9973);   //B*x = A, B*x/B = x;
    }
    return 0;
}