http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3609
Modular Inverse
Time 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
3
3 11
4 12
5 13
Sample Output
4
Not Exist
8
References
Author: WU, Zejun
Contest: The 9th Zhejiang Provincial Collegiate Programming Contest
分析:
题目要求给出a和m的值 , 求出 ax % m == 1 % m成立时的x 的最小值 , 直接x枚举到m即可。
一开始写的时候没有想到是枚举到m, 后来队友推出m。
AC代码:
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
#include<iostream>
#include<stack>
#include<map>
#include<string>
using namespace std;
int main(){
int n, x, a, m;
scanf("%d", &n);
while(n--){
bool flag = true;
scanf("%d%d", &a, &m);
for(x = ; x <= m; x++){
if((a*x)%m == %m){
flag = false;
printf("%d\n", x);
break;
}
}
if(flag){
printf("Not Exist\n");
}
}
return ;
}