Admission to Exam

题意：即求一个最小的n，满足n的欧拉函数为k
解法：由于k比较大，可以预处理sqrt（k）中的所有素数，由公式得欧拉函数值=π(pi-1)pi^(m-1），其中pi为约数，可以预处理sqrt（k）中的所有素数，然后再进行划分
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一下为队友代码

#include <cstdio>#include <map>using namespace std;map<long long,long long> a;long long prime[100010], cnt_prime;bool b[100010];long long max(long long a, long long b){    if (a < b) return b;    return a;}long long min(long long a, long long b){    if (a < b) return a;    return b;}void getprime(){    for (long long i = 2; i <= 100000; i++)    {        if (!b[i]) prime[++cnt_prime] = i;        for (long long j = 1; j <= cnt_prime && i * prime[j] <= 100000; j++)        {            b[i * prime[j]] = true;            if (i % prime[j] == 0) break;        }    }}long long count(long long p, long long pos){    if (a[p]) return a[p];    long long res = 1e18;    bool ok = false;    if (pos == 1 && p % 2 == 0)    {        long long r = 2, q = p;        while (q % 2 == 0)        {            r = r * 2; q = q / 2;            long long z = count(q, 2);            if (z < 1e18) res = min(res, r * z);        }    }    for (long long i = 1; prime[i] * prime[i] <= p + 1; i++)    {        if ((p + 1) % prime[i] == 0)        {            ok = true; break;        }    }    if (ok)    for (long long i = max(pos,2); (prime[i] - 1) * (prime[i] - 1) < p; i++)    {        if ((prime[i] - 1) * (prime[i] - 1) < p && p % (prime[i] - 1) == 0)        {            long long r = prime[i], q = p / (prime[i] - 1);             long long z = count(q, i + 1);            if (z < 1e18)            res = min(res, r * z);            while (q % prime[i] == 0)            {                r = r * prime[i]; q = q / prime[i];                z = count(q, i + 1);                if (z < 1e18)                res = min(res, r * z);            }        }    }    else res = p + 1;    a[p] = res; return res;}int main(){    long long n;    scanf("%I64d", &n);    getprime();    a[1] = 1; a[2] = 3;    long long ans = count(n, 1);    if (ans < 1e18) printf("%I64d\n", ans);    else printf("0\n");}