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2016 ACM/ICPC Asia Regional Shenyang Online && hdoj5901 Count primes Lehmer
Count primes
Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Problem Description
Easy question! Calculate how many primes between [1...n]!
Input
Each line contain one integer n(1 <= n <= 1e11).Process to end of file.
Output
For each case, output the number of primes in interval [1...n]
Sample Input
2310
Sample Output
124
Source
2016 ACM/ICPC Asia Regional Shenyang Online
粘来的板子。。留着用
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 #define MAXN 110 5 #define MAXM 10010 6 #define MAXP 666666 7 #define MAX 1000010 8 #define LL long long int 9 #define clr(arr) memset(arr,0,sizeof(arr))10 #define setbit(ar, i) (((ar[(i) >> 6]) |= (1 << (((i) >> 1) & 31))))11 #define chkbit(ar, i) (((ar[(i) >> 6]) & (1 << (((i) >> 1) & 31))))12 #define isprime(x) (( (x) && ((x)&1) && (!chkbit(arr, (x)))) || ((x) == 2))13 14 LL dp[MAXN][MAXM];15 unsigned int arr[(MAX>>6)+5]={0};16 int len=0,primes[MAXP],counter[MAX];17 18 void SS(){19 setbit(arr,0);20 setbit(arr,1);21 for(int i=3;(i*i)<MAX;i++,i++)22 if (!chkbit(arr,i)){23 int k=i<<1;24 for(int j=i*i;j<MAX;j+=k)25 setbit(arr,j);26 }27 for(int i=1;i<MAX;i++){28 counter[i]=counter[i-1];29 if (isprime(i)) primes[len++]=i,counter[i]++;30 }31 }32 33 void init(){34 SS();35 for(int n=0;n<MAXN;n++)36 for(int m=0;m<MAXM;m++){37 if (!n) dp[n][m]=m;38 else dp[n][m]=dp[n-1][m]-dp[n-1][m/primes[n-1]];39 }40 }41 42 LL phi(LL m,int n){43 if (!n) return m;44 if (primes[n-1]>=m) return 1;45 if (m<MAXM&&n<MAXN) return dp[n][m];46 return phi(m,n-1)-phi(m/primes[n-1],n-1);47 }48 49 LL Lehmer(LL m){50 if (m<MAX) return counter[m];51 int s=sqrt(0.9+m);52 int y=cbrt(0.9+m);53 int a=counter[y];54 LL res=phi(m,a)+a-1;55 for(int i=a;primes[i]<=s;i++)56 res=res-Lehmer(m/primes[i])+Lehmer(primes[i])-1;57 return res;58 }59 60 int main(){61 init();62 LL n;63 while(scanf("%I64d",&n)!=EOF) printf("%I64d\n",Lehmer(n));64 return 0;65 }
2016 ACM/ICPC Asia Regional Shenyang Online && hdoj5901 Count primes Lehmer
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