HDU 6623 Minimal Power of Prime
Time limit 1000 ms
Memory limit 65536 kB
OS Windows
中文题意
给一个数n,设将n质因数分解后可以得到
其中\(\omega(n)\)意思是n的不同质因子个数,\(a_i\)是n的质因子。
要求输出最小的\(p_i\)。
题解
看完题解感觉很妙啊——
Let's first factorize \(N\) using prime numbers not larger than \(N^{\frac{1}{5}}\). And let's denote \(M\) as the left part, all the prime factors of \(M\) are larger than \(N^{\frac{1}{5}}\). If \(M=1\) then we are done, otherwise M can only be \(P^2\), \(P^3\), \(P^4\) or \(P^2 \times Q^2\), here \(P\) and \(Q\) are prime numbers.
- If \(M^{\frac{1}{4}}\) is an integer, we can know that \(M=P^4\). Update answer using 4 and return.
- If \(M^{\frac{1}{3}}\) is an integer, we can know that \(M=P^3\). Update answer using 3 and return.
- If \(M^{\frac{1}{2}}\) is an integer, we can know that \(M=P^2\) or \(M=P^2 \times Q^2\). No matter which situation, we can always update answer using 2 and return.
- If (1)(2)(3) are false, we can know that answer=1.
Since there are just \(O(\frac{N^{\frac{1}{5}}}{log(N)})\) prime numbers, so the expected running time is \(O(\frac{TN^{\frac{1}{5}}}{log(N)})\).
官方题解导致我一开始没反应过来的地方在于
M can only be \(P^2\), \(P^3\), \(P^4\) or \(P^2 \times Q^2\),
不是only,\(M\)还可以是\(PQRS\)、\(P^2QR\)、\(PQR\)、\(P^3Q\)、\(P^2Q\),之类的,而这些的答案都是1,也就是题解里编号4所说的,不是前三种情况。
另外一个问题,如何判断\(\sqrt[4]M\)、\(\sqrt[3]M\)、\(\sqrt{M}\)是否是整数呢?我们可以求出\(\sqrt[4]M\)、\(\sqrt[3]M\)、\(\sqrt{M}\)向下取整后的结果,再乘回去,比如,看看是否存在\(\lfloor\sqrt{M}\rfloor^2==M\)。
还有,求\(\lfloor\sqrt[3]{M}\rfloor\)时,pow函数精度不太够,用pow(n,1.0/3.0)
会WA,要二分法求。下面这段二分法的代码来自这个博客代码里的two函数,要是有整数解就返回整数解,否则返回负一,这倒是挺好。
还有……复杂度那个log哪里来的?和质数分布有关?
源代码
#include<stdio.h>
#include<math.h>
#include<algorithm>
int T;
long long n;
long long cnt=0;
long long prime[10000];
long long ans;
bool vis[10000];
void shai()//取变量名一次回到解放前。还别说,挺方便的,一目了然
{
for(int i=2;i<=10000;i++)
{
if(!vis[i]) prime[++cnt]=i;
for(int j=1;j<=cnt&&i*prime[j]<=10000;j++)
{
vis[i*prime[j]]=1;
if(i%prime[j]==0)
break;
}
}
}
long long sancigeng(long long a)
{
long long l=1,r=(long long)pow(n*1.0, 1.0 / 3) + 1,mid;
while(l<=r){
mid=(l+r)>>1;
if(mid*mid*mid==n) return mid;
else if(mid*mid*mid>n) r=mid-1;
else l=mid+1;
}
return -1;
}
int main()
{
shai();
//freopen("test.in","r",stdin);
scanf("%d",&T);
while(T--)
{
scanf("%lld",&n);
ans=0x7fffffff;
int pw,pr;
for(int i=1;i<=cnt;i++)
{
if(n%prime[i]==0)
{
pr=prime[i],pw=0;
while(n%pr==0)
{
n/=pr;
pw++;
}
if(pw<ans) ans=pw;
}
if(ans==1)
{
n=1;
break;
}
}
if(n==1)
{
printf("%lld\n",ans);
continue;
}
long long temp1=sqrt(n),temp2=sqrt(temp1),temp3=sancigeng(n);
if(temp2*temp2==temp1&&temp1*temp1==n)
{//algorithm里的max和min不支持long long和int混杂的参数,居然不会隐式类型转换
ans=std::min(ans,4LL);
}
else if(temp3>=0)
{
ans=std::min(ans,3LL);
}
else if(temp1*temp1==n)
{
ans=std::min(ans,2LL);
}
else ans=1;
printf("%lld\n",ans);
}
return 0;
}