[51NOD1181]质数中的质数（质数筛法）（欧拉筛）

题目链接：http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1181

思路：欧拉筛出所有素数和一个数的判定，找到大于n的最小质数序号p，并且判断p是不是质数，输出这个数。

     /*
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 ┛┗┛┗┛┃ノ)
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 */
 #include <algorithm>
 #include <iostream>
 #include <iomanip>
 #include <cstring>
 #include <climits>
 #include <complex>
 #include <fstream>
 #include <cassert>
 #include <cstdio>
 #include <bitset>
 #include <vector>
 #include <deque>
 #include <queue>
 #include <stack>
 #include <ctime>
 #include <set>
 #include <map>
 #include <cmath>
 using namespace std;
 #define fr first
 #define sc second
 #define cl clear
 #define BUG puts("here!!!")
 #define W(a) while(a--)
 #define pb(a) push_back(a)
 #define Rint(a) scanf("%d", &a)
 #define Rll(a) scanf("%lld", &a)
 #define Rs(a) scanf("%s", a)
 #define Cin(a) cin >> a
 #define FRead() freopen("in", "r", stdin)
 #define FWrite() freopen("out", "w", stdout)
 #define Rep(i, len) for(int i = 0; i < (len); i++)
 #define For(i, a, len) for(int i = (a); i < (len); i++)
 #define Cls(a) memset((a), 0, sizeof(a))
 #define Clr(a, x) memset((a), (x), sizeof(a))
 #define Full(a) memset((a), 0x7f7f7f, sizeof(a))
 #define lrt rt << 1
 #define rrt rt << 1 | 1
 #define pi 3.14159265359
 #define RT return
 #define lowbit(x) x & (-x)
 #define onenum(x) __builtin_popcount(x)
 typedef long long LL;
 typedef long double LD;
 typedef unsigned long long ULL;
 typedef pair<int, int> pii;
 typedef pair<string, int> psi;
 typedef pair<LL, LL> pll;
 typedef map<string, int> msi;
 typedef vector<int> vi;
 typedef vector<LL> vl;
 typedef vector<vl> vvl;
 typedef vector<bool> vb;

 const int maxn = ;
 int n, pcnt;
 bool isprime[maxn];
 int prime[maxn];

 int main() {
     // FRead();
     Cls(prime); Clr(isprime, true); pcnt = ;
     For(i, , maxn) {
         if(isprime[i]) prime[++pcnt] = i;
         For(j, , pcnt+) {
             if(i * prime[j] > maxn) break;
             isprime[i*prime[j]] = ;
             if(i % prime[j] == ) break;
         }
     }
     while(~Rint(n)) {
         int p = ;
         while(prime[p] < n) p++;
         while() {
             if(isprime[p]) break;
             p++;
         }
         printf("%d\n", prime[p]);
     }
     RT ;
 }