HDU 1576 A/B 【带简单处理的扩展欧几里得】

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1576

求解思路：从A一定是B的倍数入手，令A=k*9973+C,那么k*9973+C=tB，所以tB-k*9973=C，用扩展欧几里得求出最小的正t即是A/B，那么再去一次模就求得了解

AC代码：

//HDU-1576 A/B
//AC 2016-4-19 19:54:46
//extent Euclid
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cctype>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <set>
#include <string>
#include <map>
#include <queue>
#include <deque>
#include <list>
#include <sstream>
#include <stack>
using namespace std;

#define cls(x) memset(x,0,sizeof x)
#define inf(x) memset(x,0x3f,sizeof x)
#define neg(x) memset(x,-1,sizeof x)
#define ninf(x) memset(x,0xc0,sizeof x)
#define st0(x) memset(x,false,sizeof x)
#define st1(x) memset(x,true,sizeof x)
#define INF 0x3f3f3f3f
#define lowbit(x) x&(-x)
#define abs(x) (x>0?x:-x)
#define max(x,y) (x>y?x:y)
#define min(x,y) (x<y?x:y)
#define bug cout<<"here"<<endl;
//#define debug

long long X,Y;
long long extgcd(int a,int b)//return gcd, X&Y are global
{
    if(b==0)
    {
        X=1;
        Y=0;
        return a;
    }
    long long ans=extgcd(b,a%b);
    long long temp=Y;
    Y=X-(a/b)*Y;
    X=temp;
    return ans;
}

int main()
{
    #ifdef debug
        freopen("E:\Documents\code\input.txt","r",stdin);
        freopen("E:\Documents\code\output.txt","w",stdout);
    #endif
    long long A,B;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%lld %lld",&A,&B);
        long long gcd=extgcd(B,-9973);
        long long t=A/gcd;
        long long s=abs(9973/gcd);
        long long res=((X%s*t%s)%s+s)%s;
        printf("%lld
",res);
    }
    return 0;
}