HDU 3370 Description has only two Sentences (欧拉函数+欧拉定理)
欧拉定理表明,若n,a为正整数,且n,a互质,则: ![](https://gss2.bdstatic.com/9fo3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D116/sign=b4a8aa2a7dd98d1072d40830173eb807/0823dd54564e92584b7cba389d82d158cdbf4e9f.jpg)
![](https://gss2.bdstatic.com/9fo3dSag_xI4khGkpoWK1HF6hhy/baike/s%3D116/sign=b4a8aa2a7dd98d1072d40830173eb807/0823dd54564e92584b7cba389d82d158cdbf4e9f.jpg)
费马小定理:
a是不能被质数p整除的正整数,则有a^(p-1) ≡ 1 (mod p)
证明这个定理非常简单,由于p是质数,所以有φ(p) = p-1,代入欧拉定理即可证明。推论:对于任意正整数a,有a^p ≡ a (mod p),因为a能被p整除时结论显然成立。
Description has only two Sentences
a n = X*a n-1 + Y and Y mod (X-1) = 0.
Your task is to calculate the smallest positive integer k that a k mod a 0 = 0.
Input Each line will contain only three integers X, Y, a 0 ( 1 < X < 2 31, 0 <= Y < 2 63, 0 < a 0 < 2 31).OutputFor each case, output the answer in one line, if there is no such k, output "Impossible!".Your task is to calculate the smallest positive integer k that a k mod a 0 = 0.
Sample Input
2 0 9Sample Output
1
题意: 给出公式及x,y,a0的值(y%(x-1)=0),求使ak%a0=0的k的最小值。
思路:在纸上推一下可以推出所求的k值是使(x^k-1)*(y/(x-1))=0(mod a0)。令B=(y/(x-1)),为了使用欧拉定理,必须把(y/(x-1))消去,所以求gcd(a0,B),且a0=a0/gcd。
所以,问题转化为求最小的k使x^k=1(mod a0),而欧拉定理要求x与a0互质。所以,首先判断gcd(x,a0)是否为1:若不为1,则无解;否则用欧拉定理求解。
(先求出a0的欧拉函数值,然后枚举k的取值,注意用i*i<=phi(a0),不然会超时。)
AC代码:
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#define LL long long
using namespace std;
LL euler(LL n)
{
LL res=n,a=n;
for(int i=2;i*i<a;i++)
if(a%i==0)
{ res=res/i*(i-1);
while(a%i==0) a/=i; }
if(a>1) res=res/a*(a-1);
return res;
}
LL gcd(LL a,LL b)
{ if(b==0) return a; else return gcd(b,a%b); }
LL ppow(LL a,LL n,LL mod)
{
LL ans=1;
while(n) { if(n%2==1) ans=(ans*a)%mod; a=(a*a)%mod; n/=2; }
return ans%mod;
}
int main()
{
LL x,y,a0;
while(scanf("%lld%lld%lld",&x,&y,&a0)!=EOF) {
LL c=y/(x-1);
LL g=gcd(c,a0);
a0/=g;
if(gcd(x,a0)!=1) {
cout<<"Impossible!\n";
continue;
}
LL n=euler(a0);
LL m=n;
LL i;
LL ans=n;
for(i=2;i*i<=m;i++)
{
if(m%i==0)
{
if(ppow(x,i,a0)==1) ans=min(ans,i);
if(ppow(x,m/i,a0)==1) ans=min(ans,m/i);
}
}
cout<<ans<<endl;
}
return 0;
}