HDU 1576 A/B

Description

要求(A/B)%9973，但由于A很大，我们只给出n(n=A%9973)(我们给定的A必能被B整除，且gcd(B,9973) = 1)。

Input

数据的第一行是一个T，表示有T组数据。 

每组数据有两个数n(0 <= n < 9973)和B(1 <= B <= 10^9)。

Output

对应每组数据输出(A/B)%9973。

Sample Input

2
1000 53
87 123456789      

Sample Output

7922

6060

#include<set>
#include<map>
#include<ctime>
#include<cmath>
#include<stack>
#include<queue>
#include<bitset>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define rep(i,j,k) for (int i = j; i <= k; i++)
#define per(i,j,k) for (int i = j; i >= k; i--)
#define loop(i,j,k) for (int i = j;i != -1; i = k[i])
#define lson x << 1, l, mid
#define rson x << 1 | 1, mid + 1, r
#define fi first
#define se second
#define mp(i,j) make_pair(i,j)
#define pii pair<string,string>
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const double eps = 1e-8;
const int INF = 0x7FFFFFFF;
const int mod = 9973;
const int N = 5e3 + 10;
const int read()
{
  char ch = getchar();
  while (ch<'0' || ch>'9') ch = getchar();
  int x = ch - '0';
  while ((ch = getchar()) >= '0'&&ch <= '9') x = x * 10 + ch - '0';
  return x;
}
int T, n, m;

int exgcd(int a, int b, int &x, int &y)
{
  if (!b) { x = 1, y = 0; return a; }
  int g = exgcd(b, a%b, x, y);
  int z = x - a / b * y;
  x = y;  y = z;  return g;
}

int main()
{
  T = read();
  while (T--)
  {
    scanf("%d%d", &n, &m);
    int x, y;
    exgcd(m, mod, x, y);
    printf("%d\n", n * (x + mod) % mod);
  }
  return 0;
}