POJ-3641：Pseudoprime numbers（快速幂）
POJ-3641：Pseudoprime numbers（快速幂）
题目：
Fermat's theorem states that for any prime number p and for any integer a > 1, ap = a (mod p). That is, if we raise a to the pth power and divide by p, the remainder is a. Some (but not very many) non-prime values of p, known as base-a pseudoprimes, have this property for some a. (And some, known as Carmichael Numbers, are base-a pseudoprimes for all a.)
Given 2 < p ≤ 1000000000 and 1 < a < p, determine whether or not p is a base-a pseudoprime.
Input
Input contains several test cases followed by a line containing "0 0". Each test case consists of a line containing p and a.
Output
For each test case, output "yes" if p is a base-a pseudoprime; otherwise output "no".
Sample Input
3 2
10 3
341 2
341 3
1105 2
1105 3
0 0Sample Output
no no yes no yes yes
概译：费马定理觉得所有的素数p都是，给定一个大于1小于p的a， a的p次幂模p的值为a。然后有一些不是素数的p对于某些a也有同样的性质，我们就叫他基于a 的Pseudoprime（a从1到p都成立的就叫Carmichael number啦，别的题里大家应该见过）。现在给一些组p和a，判断一下是不是Pseudoprime number。
题目很水，只是借机留下几个基础的模板。详见代码：
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
bool isprime(int n)//试除法判定素数
{
int m=floor(sqrt(n)+0.5);
//这里sqrt应该都懂……+0.5是避免浮点数误差，因为下面枚举时举i<=m的，所以不要紧
//如果不用变量m而是直接把sqrt写循环里，那会导致每次都计算一遍sqrt，会增加时间
for(int i=2;i<=m;i++)
if(n%i==0)
return false;
return n>1;//最后滤去0和1
}
int qmul(int a,int b,int mod)//快速乘，类似于快速幂
{
int ans=0;
while(b)
{
if(b&1)
ans=(ans+a)%mod;
b>>=1;
a=a*2%mod;
}
return ans;
}
int qpow(int a,int n,int mod)//快速幂，自己在纸上随便列两个数写一遍就懂了
{
int res=1;
while(n)
{
if(n&1)
res=qmul(res,a,mod)%mod;//原型为res=res*a%mod，这里没用longlong怕爆
n>>=1;
a=qmul(a,a,mod)%mod;//原型为a=a*a%mod
}
return res;
}
int main()
{
int p,a;
while(~scanf("%d%d",&p,&a)&&p|a)
printf("%s\n",a==qpow(a,p,p)&&!isprime(p)?"yes":"no");
return 0;
}