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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)
.
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.
For each test case, output the smallest positive x. If such x doesn‘t exist, output "Not Exist".
3 3 11 4 12 5 13
4 Not Exist 8
#include<cstdio> int gcd(int a,int b) { return b==0?a:gcd(b,a%b); } int main() { int i,j,k; int a,m; int T; scanf("%d",&T); while(T--) { scanf("%d%d",&a,&m); if(gcd(a,m)!=1) {printf("Not Exist\n");continue;} //排位赛的时候一直在这里wa掉,谨记: //a三b(mod m) 是同余 即:a mod m == b mod m //可表示成 a=b+m*k 其中k是从 0 开始 for(k=0;k<=1000;k++) { if((m*k+1)%a==0) {printf("%d\n",(m*k+1)/a);break;} } } return 0; }
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原文地址:http://www.cnblogs.com/orchidzjl/p/4307530.html