ZOJ 3609 Modular Inverse (水题)
2014-08-21 10:32
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Modular Inverse
Time Limit: 2 Seconds Memory Limit: 65536 KBThe 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_inverseAuthor: WU, Zejun
Contest: The 9th Zhejiang Provincial Collegiate Programming Contest
简单来说就是要求给定n,m 求一个x使得 (n*x)%m=1, 如果x存在输出最小正整数x,否则输出Not Exist
注意m=1的情况,因为任何数对1取模会等于0,但是这里要求输出最小正整数,所以输出1
#include<cstdio> #include<cstring> #include<stdlib.h> #include<algorithm> using namespace std; int gcd(int a,int b) { return b?gcd(b,a%b):a; } int main() { //freopen("in.txt","r",stdin); int kase; scanf("%d",&kase); while(kase--) { int n,m; scanf("%d %d",&n,&m); if(m==1)//当m=1时,数字对1取模等于0,存在这个数字,但是这里要输出最小的正整数,所以输出1 {printf("1\n");continue;} int Gcd=gcd(n,m); if(Gcd>1) {printf("Not Exist\n");continue;} else for(int i=1;i<=1000;i++) if((n*i)%m==1) {printf("%d\n",i);break;} } return 0; }
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