ZOJ 3609 Modular Inverse (水題)

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Modular InverseTime Limit: 2 Seconds      Memory Limit: 65536 KB

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 Exist8

References

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

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

 

簡單來說就是要求給定n,m 求一個x使得 (n*x)%m=1, 如果x存在輸出最小正整數x，否則輸出Not Exist

注意m=1的情況，因為任何數對1模數會等於0，但是這裡要求輸出最小正整數，所以輸出1

 1 #include<cstdio> 2 #include<cstring> 3 #include<stdlib.h> 4 #include<algorithm> 5 using namespace std; 6 int gcd(int a,int b) 7 { 8     return b?gcd(b,a%b):a; 9 }10 int main()11 {12     //freopen("in.txt","r",stdin);13     int kase;14     scanf("%d",&kase);15     while(kase--)16     {17         int n,m;18         scanf("%d %d",&n,&m);19         20         if(m==1)//當m=1時，數字對1模數等於0，存在這個數字，但是這裡要輸出最小的正整數，所以輸出121         {printf("1\n");continue;}22         23         int Gcd=gcd(n,m);24         25         if(Gcd>1)26         {printf("Not Exist\n");continue;}27          28         else29             for(int i=1;i<=1000;i++)30                 if((n*i)%m==1)31                 {printf("%d\n",i);break;}   32     }33     return 0;34 }

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