傳送門：http://www.lydsy.com/JudgeOnline/problem.php?id=3643

【題解】

n = p1^a1*p2^a2*...*pm^am

phi(n) = p1(p1-1)^(a1-1)*p2(p2-1)^(a2-1)*...*pm^(am-1)

最多有10個不同的質因數就超過maxint了，這告訴我們可以搜索

我們假設p1<p2<p3<...<pm

那麽我們處理出<sqrt(n)的質數，因為我們只會用到這些質數來搜（因為其他平方完就爆炸了）

那麽我們就暴力嘗試是否被(pi-1)整除，是的話枚舉因子個數，dfs下去即可

註意特判如果我當前剩下了n，n+1是個質數，那麽也可以是直接當前答案*(n+1)

復雜度……我咋知道

O（能過）

# include <math.h>
# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 5e4 + 10
 
;
const int mod = 1e9+7;

# define RG register
# define ST static

ll ans = 1e18;
int n, F;
bool isnp[M];
int p[M/3], pn=0;

inline bool isprime(int n) {
    for (int i=1, to = sqrt(n); p[i] <= to; ++i)
        if(n % p[i] == 0) return false;
    return true;
}

inline void dfs(int n, int lst, ll sum) {
    
 
if(sum >= ans) return;
    if(n == 1) { ans = sum; return ; }
    if(n > F && isprime(n+1)) ans = min(ans, sum*(n+1));
    for (int i=lst+1; p[i]-1 <= F && p[i]-1 <= n; ++i) {
        if(n % (p[i]-1) == 0) {
            int t = n/(p[i]-1); ll res = sum * p[i];
            dfs(t, i, res);
            while(t % p[i] == 0) {
                t /= p[i];
                res *= p[i];
                dfs(t, i, res);
            }
        }
    }
}

int main() {
    cin >> n;
    F = sqrt(n);
    isnp[0] = isnp[1] = 1;
    for (int i=2; i<=50000; ++i) {
        if(!isnp[i]) p[++pn] = i;
        for (int j=1; j<=pn && i*p[j] <= 50000; ++j) {
            isnp[i*p[j]] = 1;
            if(i%p[j] == 0) break;
        }
    }
    dfs(n, 0, 1);
    if(ans <= 2147483647) cout << ans << endl;
    else cout << -1 << endl;
    return 0;
}

bzoj3643 Phi的反函數