J - Modular Inverse ZOJ - 3609 ----逆元

bless295

于 2017-04-28 20:50:58 发布

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本文介绍了一种求解模逆元的算法，并通过具体的输入输出样例进行说明。该算法利用扩展欧几里得算法解决整数a关于模m的逆元问题，即找到一个最小正整数x，使得ax ≡ 1 (mod m)。

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J - Modular Inverse

ZOJ - 3609

The modular modular multiplicative inverse of an integer a modulo m is an integer xsuch 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

Sample Output

4
Not Exist
8

References

http://en.wikipedia.org/wiki/Modular_inverse

题目链接：https://cn.vjudge.net/contest/160971#problem/J

直接裸求逆元就好

代码：

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
int exgcd(int a,int b,int& x,int& y){
	if(b==0){
		x=1;
		y=0;
		return a;
	}
	int r=exgcd(b,a%b,x,y);
	int t=y;
	y=x-(a/b)*y;
	x=t;
	return r;
}
int main(){
	int t;
	scanf("%d",&t);
    while(t--){
        int a,m;
        int x,y;
        scanf("%d%d",&a,&m);
        int res=exgcd(a,m,x,y);
        if(res!=1){
            printf("Not Exist\n");
            continue;
        }
        int ans=x;
        ans=ans%m;
        if(ans<=0) ans+=m;
        printf("%d\n",ans);
    }
}