ZOJ

Modular Inverse

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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

33 114 125 13

Sample Output

4Not Exist
8
分析：题目让求a相对于m的逆元 x
从题目可以知道  gcd(a,m)=1;
扩展欧几里德  a*x+m*y=gcd(a,m)
求出x
AC代码：
#include<stdio.h>#include<string.h>long long kzgcd(long long a,long long b,long long &x,long long &y){if(b==0){x=1,y=0;return a;}long long ans=kzgcd(b,a%b,x,y);long long t=x;x=y;y=t-a/b*y;return ans;}long long solve(long long a,long long b){long long x,y;long long gcd=kzgcd(a,b,x,y);if(1%gcd)return -1;b/=gcd;if(b<0) b=-b;long long ans=x%b;if(ans<=0)ans+=b;return ans;}int main(){int T;scanf("%d",&T);while(T--){long long a,m;scanf("%lld%lld",&a,&m);long long ans=solve(a,m);if(ans==-1)printf("Not Exist\n");elseprintf("%lld\n",ans);} }