ZOJ-3609(Modular Inverse ）(拓展欧几里得)

The modular modular multiplicative inverse of an integer a modulo m is an integer x such that a-1≡x (mod m). This is equivalent to ax≡1 (mod m).

Input

There are multiple test cases. The first line of input is an integer T ≈ 2000 indicating the number of test cases.

Each test case contains two integers 0 < a ≤ 1000 and 0 < m ≤ 1000.

Output

For each test case, output the smallest positive x. If such x doesn’t exist, output “Not Exist”.

Sample Input

3

3 11

4 12

5 13

Sample Output

4

Not Exist

8

References

题意，求一个最小的正数使得ax %m == 1

我们可以将这个式子展开即(ax-1) % m = 0 ,即ax-1 = y*m,将其移动到右边即是ax - my = 1,显然我们要用到扩展欧几里得，详细解释在代码注释里面

代码如下:

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
int t,a,m;
typedef long long ll;
ll extgcd(ll a,ll b,ll &x,ll &y){//我表示没有看懂，展示当模板强行记住，我好笨啊。
ll d = a;
if(b != 0){
d = extgcd(b,a%b,y,x);
y -= (a/b)*x;
}else{
x = 1;
y = 0;
}
return d;
}
int main(void)
{
ll t;
cin >> t;
while(t--){
cin >> a >> m;
ll x,y;
ll ans = extgcd(a,m,x,y);
if(ans != 1)
cout << "Not Exist" << endl;
else{
/*因为拓展欧几里得是求ax+by=1的解，这里是ax-by=1,但当求的y为负数时 也可默认为前着转化为了后者
例如，我们得到一组解x0,y0那么a(x0+b)+b(y0-a)与原式相等，当x为正时，则y一定为负.*/
while(x <= 0)
x += m;
cout << x << endl;
}
}
return 0;
}