Modular inverse (modulo inverse element, extended Euclidean)

Modular Inverse Time limit: 2 Seconds Memory Limit: 65536 KB

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

Input

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

Each test case contains 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

33 114 125 13

Sample Output

4Not Exist8
Solution: To find the smallest positive integer, in fact, the general solution of X is x0+b/gcd*t, because T is an integer, then ans=x0+b/gcd*t=x0 mod b=x0%b; because ans, if it's a positive integer,
So when the B/GCD is negative, it is equal to the absolute value, because there is t ah, when the x0%b negative, plus a B; it is, because ans= (x0+b)%b;
Code:

1#include <iostream>2#include <algorithm>3#include <cstdio>4#include <cstring>5#include <cmath>6 using namespacestd;7 Const intinf=0x3f3f3f3f;8typedefLong LongLL;9 voidE_GCD (LL a,ll b,ll &d,ll &x,ll &y) {Ten     if(!b) { OneD=A; Ax=1; -y=0; -     } the     Else{ -E_GCD (b,a%b,d,x,y); -LL temp=x; -x=y; +y=temp-a/b*y; -     } + } ALL Cal (intAintBintc) { at LL x,y,d; - E_GCD (a,b,d,x,y); -     if(c%d!=0)return-1;//ax+by=c/(C/GCD); -x*=c/D; -B/=d;//because the general solution of X is x0+ (B/GCD) t; -      if(b<0) b=-b; inLL ans=x%b; -      if(ans<=0) ans+=b; to      returnans; + } - intMain () { the LL t,a,b,d,x,y; *scanf"%d",&T); $      while(t--){Panax Notoginsengscanf"%lld%lld",&a,&b); -LL Ans=cal (A, B,1); the         if(ans==-1) puts ("Not Exist"); +         Elseprintf"%lld\n", ans); A     } the     return 0; +}

Data on the problem comparison of water, data range 1000, violence can be:

#include <iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<cmath>using namespacestd;Const intinf=0x3f3f3f3f; typedefLong LongLL;intMain () {intt,a,m; scanf ("%d",&T);  while(t--) {//(1-ax)%m;scanf"%d%d",&a,&m); intflot=0;  for(intx=1; x<= +; x + +){            if((1-A*X)%m==0) {Flot=1; printf ("%d\n", x);  Break; }        }        if(Flot)Continue; Puts ("Not Exist"); }    return 0;}

Modular inverse (modulo inverse element, extended Euclidean)