HDU 1431 Prime Number

HDU 1431 Prime Number

Someone asked me this question.

I personally feel that the violent search will be TLE O (n * sqrt (n )). N = 100000000; (2 ~ is used to determine the prime number ~ Sqrt (n) + 1 removal)

The enumeration is better. Enumeration 1 ~ 10000. Save each of them. The number of input records is known as left, right, and then combine them.

For example, 1, judge whether 11 is a prime number.

For example, 10 is used to determine whether 101 is a prime number and whether 1001 is a prime number.

The complexity is O (n ^ 2 ). At the beginning, bool pa [100000000] is ready to be identified with tags. Result MLE.

Then, calculate the total number, which can be 781 at most. Int pa [1000] can be mounted.

G ++ 15 ms

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            #define INF 0x7fffffff#define eps 1e-8#define LL long long#define PI 3.141592654#define CLR(a,b) memset(a,b,sizeof(a))#define FOR(i,a,b) for(int i=a;i
            
             =b;i--)#define pb push_back#define mp make_pair#define ft first#define sd second#define sf scanf#define pf printf#define sz(v) ((int)(v).size())#define all(v) (v).begin(),(v).end()#define acfun std::ios::sync_with_stdio(false)#define SIZE 100000000 +1using namespace std;int pa[1000];int cot;bool prime(int n){ FOR(i,2,sqrt(n)+2) if(n%i==0)return 0; return 1;}void PA(){ cot=0; pa[cot++]=2; pa[cot++]=3; pa[cot++]=5; pa[cot++]=7; FOR(i,1,10000) { int num[5]; int len=0; int m=i; while(m) { int tmp=m%10; num[len++]=tmp; m/=10; } int ans=i; if(len>1) { FOR(r,1,len) ans=ans*10+num[r]; if(prime(ans)) pa[cot++]=ans; } ans=i; FOR(r,0,len) ans=ans*10+num[r]; if(prime(ans)) pa[cot++]=ans; }}int main(){ PA(); int a,b; while(~sf("%d%d",&a,&b)) { FOR(i,0,cot) if(pa[i]>=a&&pa[i]<=b)pf("%d\n",pa[i]); pf("\n"); }}