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
題解：求最小正整數解，其實吧，x的通解是x0+b/gcd*t，由於t是整數，那么ans=x0+b/gcd*t=x0 mod b=x0%b；因為ans要是正整數的，
所以當b/gcd是負的時候，就等於絕對值就好了，因為還有t啊，當x0%b負時,加上一個b；就妥了；因為ans=(x0+b)%b;
代碼：
1 #include<iostream> 2 #include<algorithm> 3 #include<cstdio> 4 #include<cstring> 5 #include<cmath> 6 using namespace std; 7 const int INF=0x3f3f3f3f; 8 typedef long long LL; 9 void e_gcd(LL a,LL b,LL &d,LL &x,LL &y){ 10 if(!b){ 11 d=a; 12 x=1; 13 y=0; 14 } 15 else{ 16 e_gcd(b,a%b,d,x,y); 17 LL temp=x; 18 x=y; 19 y=temp-a/b*y; 20 } 21 } 22 LL cal(int a,int b,int c){ 23 LL x,y,d; 24 e_gcd(a,b,d,x,y); 25 if(c%d!=0)return -1;//ax+by=c/(c/gcd); 26 x*=c/d; 27 b/=d;//因為x的通解是x0+(b/gcd)t; 28 if(b<0)b=-b; 29 LL ans=x%b; 30 if(ans<=0)ans+=b; 31 return ans; 32 } 33 int main(){ 34 LL T,a,b,d,x,y; 35 scanf("%d",&T); 36 while(T--){ 37 scanf("%lld%lld",&a,&b); 38 LL ans=cal(a,b,1); 39 if(ans==-1)puts("Not Exist"); 40 else printf("%lld\n",ans); 41 } 42 return 0; 43 }
題上數據比較水，數據范圍1000，暴力一下就可以了：
#include<iostream> #include<algorithm> #include<cstdio> #include<cstring> #include<cmath> using namespace std; const int INF=0x3f3f3f3f; typedef long long LL; int main(){ int T,a,m; scanf("%d",&T); while(T--){//(1-ax)%m; scanf("%d%d",&a,&m); int flot=0; for(int x=1;x<=1000;x++){ if((1-a*x)%m==0){ flot=1; printf("%d\n",x); break; } } if(flot)continue; puts("Not Exist"); } return 0; }