Title Link: http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1181
Idea: Euler sift out all primes and a number of judgements, find the smallest prime number greater than n p, and determine if p is prime, output this number.
1 /*2 ━━━━━┒ギリギリ♂eye! 3 ┓┏┓┏┓┃キリキリ♂mind! 4 ┛┗┛┗┛┃\0/5 ┓┏┓┏┓┃/6 ┛┗┛┗┛┃ノ)7 ┓┏┓┏┓┃8 ┛┗┛┗┛┃9 ┓┏┓┏┓┃Ten ┛┗┛┗┛┃ One ┓┏┓┏┓┃ A ┛┗┛┗┛┃ - ┓┏┓┏┓┃ - ┃┃┃┃┃┃ the ┻┻┻┻┻┻ - */ -#include <algorithm> -#include <iostream> +#include <iomanip> -#include <cstring> +#include <climits> A#include <complex> at#include <fstream> -#include <cassert> -#include <cstdio> -#include <bitset> -#include <vector> -#include <deque> in#include <queue> -#include <stack> to#include <ctime> +#include <Set> -#include <map> the#include <cmath> * using namespacestd; $ #defineFr FirstPanax Notoginseng #defineSC Second - #defineCL Clear the #defineBUG puts ("Here!!!") + #defineW (a) while (a--) A #definePB (a) push_back (a) the #defineRint (a) scanf ("%d", &a) + #defineRll (a) scanf ("%lld", &a) - #defineRs (a) scanf ("%s", a) $ #defineCIN (a) CIN >> a $ #defineFRead () freopen ("in", "R", stdin) - #defineFWrite () freopen ("Out", "w", stdout) - #defineRep (i, Len) for (int i = 0; i < (len); i++) the #defineFor (I, A, Len) for (int i = (a); I < (len); i++) - #defineCls (a) memset ((a), 0, sizeof (a))Wuyi #defineCLR (A, X) memset ((a), (x), sizeof (a)) the #defineFull (a) memset ((a), 0x7f7f7f, sizeof (a)) - #defineLRT RT << 1 Wu #defineRRT RT << 1 | 1 - #definePi 3.14159265359 About #defineRT return $ #defineLowbit (x) x & (-X) - #defineOnenum (x) __builtin_popcount (x) -typedefLong LongLL; -typedefLong DoubleLD; Atypedef unsignedLong LongULL; +typedef pair<int,int>PII; thetypedef pair<string,int>psi; -typedef PAIR<LL, Ll>PLL; $typedef map<string,int>MSI; thetypedef vector<int>VI; thetypedef vector<ll>VL; thetypedef vector<vl>VVL; thetypedef vector<BOOL>vb; - in Const intMAXN =1000010; the intN, pcnt; the BOOLISPRIME[MAXN]; About intPRIME[MAXN]; the the intMain () { the //FRead (); +Cls (prime); CLR (IsPrime,true); pcnt =0; -for (I,2, MAXN) { the if(Isprime[i]) prime[++pcnt] =i;BayiFor (J,1, pcnt+1) { the if(i * prime[j] > MAXN) Break; theISPRIME[I*PRIME[J]] =0; - if(i% prime[j] = =0) Break; - } the } the while(~Rint (n)) { the intp =1; the while(Prime[p] < n) p++; - while(1) { the if(Isprime[p]) Break; thep++; the }94printf"%d\n", Prime[p]); the } theRt0; the}
[51nod1181] prime number in prime (prime number Sieve method) (Euler sieve)