BJOI2014 Euler

Description

已知$\varphi (x)=y$，求最小的$x$

Solution

$$\left\{
\begin{aligned}
\varphi(x)=x\prod_{i=1}^n \frac{p_i-1}{p_i} \\
x=p_1^{q_1}p_2^{q_2}\cdots p_n^{q_n}
\end{aligned}
\right.$$

化简得

$$y=p_1^{q_1-1}p_2^{q_2-1}\cdots p_n^{q_n-1}\prod_{i=1}^n {p_i-1}$$

枚举$y$中所有因数，找到所有的$p$，DFS组合所有$x$

#pragma GCC optimize(2)
#include<algorithm>
#include<iostream>
#include<cstdio>
using namespace std;
long long T,y,ans,tot,p[1000000];
inline long long read()
{
    long long f=1,w=0;
    char ch=0;
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')
            f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        w=(w<<1)+(w<<3)+ch-'0';
        ch=getchar();
    }
    return f*w;
}
bool check(long long x)
{
    for(long long i=2;i*i<=x;i++)
        if(!(x%i))
            return false;
    return true;
}
bool split(long long cnt,long long now)
{
    while(!(now%p[cnt]))
        now/=p[cnt];
    return now==1;
}
void dfs(long long cnt,long long now,long long ret)
{
    if(cnt>tot||ret>ans)
        return;
    dfs(cnt+1,now,ret);
    if(!(now%(p[cnt]-1)))
    {
        if(split(cnt,now/(p[cnt]-1)))
            ans=min(ans,ret/(p[cnt]-1)*p[cnt]);
        now/=(p[cnt]-1);
        dfs(cnt+1,now,ret/(p[cnt]-1)*p[cnt]);
        while(!(now%p[cnt]))
        {
            now/=p[cnt];
            dfs(cnt+1,now,ret/(p[cnt]-1)*p[cnt]);
        }
    }
}
bool cmp(long long a,long long b)
{
    return a>b;
}
int main()
{
    T=read();
    for(;T;T--)
    {
        tot=0;
        y=read();
        if(y==1)
        {
            puts("1");
            continue;
        }
        ans=1ll<<60;
        for(long long i=1;i*i<=y;i++)
            if(!(y%i))
            {
                if(check(i+1ll))
                    p[++tot]=i+1ll;
                if(i*i==y)
                    continue;
                if(check(y/i+1ll))
                    p[++tot]=y/i+1ll;
            }
        sort(p+1,p+tot+1,cmp);
        dfs(1,y,y);
        printf("%lld\n",ans);
    }
    return 0;
}