BZOJ：2242: [SDOI2011]计算器

题解：BSGS

问题：map空间 

　　BSGS判无解 a%p!=0
　　0与最小非负整数 有区别
　　函数传参类型转换int->long long long long ->int;
　　费马小定理充分必要 性?

#include<iostream>
#include<cstdio>
#include<cstring>
#include<map>
#include<cmath>
using namespace std;
typedef long long Lint;

int T,k;
Lint mm;

map<Lint,int>ma;

Lint ksm(Lint a,Lint p){
	Lint ret=1;
	for(;p;p>>=1,a=a*a%mm){
		if(p&1)ret=ret*a%mm;
	}
	return ret;
}

void gcd(Lint a,Lint b,Lint &d,Lint &x,Lint &y){
	if(b==0){
		d=a;x=1;y=0;
	}else{
		gcd(b,a%b,d,y,x);
		y-=x*(a/b);
	}
}

int main(){
	scanf("%d%d",&T,&k);
	while(T--){
		if(k==1){
			Lint a,p;
			scanf("%lld%lld%lld",&a,&p,&mm);
			printf("%lld
",ksm(a,p));
		}
		if(k==2){
			Lint a,c,b,x0,y0,d;
			scanf("%lld%lld%lld",&a,&c,&b);
			gcd(a,b,d,x0,y0);
			if(c%d!=0){
				printf("Orz, I cannot find x!
");
				continue;
			}else{
				x0=c/d*x0;b=b/d;
				printf("%lld
",(x0%b+b)%b);
			}
		}
		if(k==3){
			Lint a,c,p;
			scanf("%lld%lld%lld",&a,&c,&p);
			Lint m=ceil(sqrt(p));
			ma.clear();
			int flag=0;mm=p;
			Lint tmp=ksm(a,m);
			if(a%p==0){
				printf("Orz, I cannot find x!
");
				continue;
			}
			for(Lint x=c%p,i=0;i<=m;++i,x=x*a%p)ma[x]=i;
			for(Lint x=tmp%p,i=1;i<=m;++i,x=x*tmp%p){
				if(ma.count(x)&&(m*i-ma[x]!=0)){
					printf("%d
",m*i-ma[x]);
					flag=1;
					break;
				}
			}
			if(!flag)printf("Orz, I cannot find x!
");
		}
	}
	return 0;
}