ZOJ——3609 Modular Inverse

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

3
3 11
4 12
5 13

Sample Output

4
Not Exist
8

References

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

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Author: WU, Zejun
Contest: The 9th Zhejiang Provincial Collegiate Programming Contest
Submit    Status

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Copyright @ 2001-2017, Zhejiang University ACM/ICPC Team, All rights reserved.

思路：

  裸地乘法逆元

但是让我做的可谓是错误百出。。。。（交了差不多10遍吧。。。）

错误：

1.多组数据

2。最小正整数，所以当求出来的x为0时也要加abs（b）

3.最小 那就%b

代码：

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int t,a,b,x,y,gcd;
int read()
{
    ,f=; char ch=getchar();
    ; ch=getchar();}
    +ch-'; ch=getchar();}
    return x*f;
}
int exgcd(int a,int b,int &x1,int &y1)
{
    )
    {
        x=,y=;
        return a;
    }
    int r=exgcd(b,a%b,x1,y1),tmp;
    tmp=x1,x1=y1,y1=tmp-a/b*y1;
    return r;
}
int main()
{
    while(scanf("%d",&t)!=EOF)
    while(t--)
    {
        a=read(),b=read();
        gcd=exgcd(a,b,x,y);
        x%=b;
        ) printf("Not Exist\n");
        else  {while(x<=0) x+=abs(b); printf("%d\n",x);}
    }
    ;
}