20201030 day50 复习13：逆元、裴蜀定理

逆元

纯属复习

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <stack>
using namespace std;
#define f(i,x,y) for(int (i)=x;(i)<=(y);(i)++)
#define fp(i,x,y) for(int (i)=x;(i)>=(y);(i)--)
long long read(){
	long long op=1,a=0;char c=getchar();
	while(c<'0'||c>'9') {if(c=='-') op=-1;c=getchar();}
	while(c>='0'&&c<='9') {a*=10,a+=c^48,c=getchar();}
	return a*op;
}
const int maxn=3e6+10;
#define int long long
int inv[maxn],mod,n;
//线性求逆元 
void init_xianxing(){
	inv[1]=1;
	f(i,2,n) inv[i]=(long long)(mod-mod/i)*inv[mod%i]%mod;
	return ;
}
//exgcd 
void exgcd(int a,int b,int &x,int &y){
	if(!b) x=1,y=0;
	else exgcd(b,a%b,y,x),y-=a/b*x;	
}
int get_exgcd_way(int a,int mod){
	int x,y;
	exgcd(a,mod,x,y);
	x=(x%mod+mod)%mod;
	return x;
}
//费马小定理
int fpm_get(int x,int p,int mod){
	x%=mod;
	int ans=1;
	while(p){
		if(p&1) (ans*=x)%=mod;
		p>>=1;(x*=x)%=mod;	
	}
	return ans;
}
#undef int 
int main(){
	#define int long long
	n=read(),mod=read();
//	init_xianxing();
//	f(i,1,n) printf("%lld\n",inv[i]); 
//	f(i,1,n) printf("%lld\n",get_exgcd_way(i,mod));
	f(i,1,n) printf("%lld\n",fpm_get(i,mod-2,mod));
	return 0;
}

裴蜀定理

关于\(x,y\)的不定方程\(ax + by = c\)有整数解的充要条件是\(\gcd(a, b)\mid c\)

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <stack>
using namespace std;
#define f(i,x,y) for(int (i)=x;(i)<=(y);(i)++)
#define fp(i,x,y) for(int (i)=x;(i)>=(y);(i)--)
long long read(){
	long long op=1,a=0;char c=getchar();
	while(c<'0'||c>'9') {if(c=='-') op=-1;c=getchar();}
	while(c>='0'&&c<='9') {a*=10,a+=c^48,c=getchar();}
	return a*op;
}
const int maxn=25;
#define int long long
int n,ans;
int a[maxn];
int gcd(int x,int y){return !y?x:gcd(y,x%y);}
#undef int
int main(){
	#define int long long
	n=read();
	f(i,1,n) {a[i]=read();if(a[i]<0) a[i]=-a[i];ans=gcd(ans,a[i]);}
	printf("%lld",ans);
	return 0;
}