ZOJ3609——数论基础 扩展欧几里得求解乘法逆元

原题如下：

Description

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 toax≡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 Exist8

题目大意：裸的求模的乘法逆元的题目，要注意得有要是正解。所以若解为负的，得加上解得间隔直到解为正数。间隔为m/gcd(a,m)。

代码如下

#include <stdio.h>#include <algorithm>#include <string.h>#include <cctype>#include <queue>using namespace std;

//欧几里得算法int gcd(int a,int b){    if(b==0)        return a;    return gcd(b,a%b);}

//扩展欧几里得算法int exgcd(int a,int b,int &x,int &y){    int d=a;    if(b!=0)    {        d=exgcd(b,a%b,y,x);        y-=(a/b)*x;    }    else    {        x=1;        y=0;    }    return d;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        int a,m;        scanf("%d%d",&a,&m);        if(gcd(a,m)!=1)//如果最大公因数不是1，则表明无解        {            printf("Not Exist\n");            continue;        }        int x,y;        exgcd(a,m,x,y);        while(x<=0)        {            x+=m/gcd(a,m);//使结果为正数        }        printf("%d\n",x);    }}