题目链接:点击打开链接
题目意思就是求1到n之间的素数个数 n<=1e11。
思路就是: 不会啊, 先存两份模板.
顺便存存高手代码。暂时没看懂,不过好短,感觉很厉害。
#include <bits/stdtr1c++.h>
#define MAXN 100
#define MAXM 10001
#define MAXP 40000
#define MAX 400000
#define clr(ar) memset(ar, 0, sizeof(ar))
#define read() freopen("lol.txt", "r", stdin)
#define dbg(x) cout << #x << " = " << x << endl
#define chkbit(ar, i) (((ar[(i) >> 6]) & (1 << (((i) >> 1) & 31))))
#define setbit(ar, i) (((ar[(i) >> 6]) |= (1 << (((i) >> 1) & 31))))
#define isprime(x) (( (x) && ((x)&1) && (!chkbit(ar, (x)))) || ((x) == 2))
using namespace std;
namespace pcf
{
long long dp[MAXN][MAXM];
unsigned int ar[(MAX >> 6) + 5] = {0};
int len = 0, primes[MAXP], counter[MAX];
void Sieve()
{
setbit(ar, 0), setbit(ar, 1);
for (int i = 3; (i * i) < MAX; i++, i++)
{
if (!chkbit(ar, i))
{
int k = i << 1;
for (int j = (i * i); j < MAX; j += k) setbit(ar, j);
}
}
for (int i = 1; i < MAX; i++)
{
counter[i] = counter[i - 1];
if (isprime(i)) primes[len++] = i, counter[i]++;
}
}
void init()
{
Sieve();
for (int n = 0; n < MAXN; n++)
{
for (int m = 0; m < MAXM; m++)
{
if (!n) dp[n][m] = m;
else dp[n][m] = dp[n - 1][m] - dp[n - 1][m / primes[n - 1]];
}
}
}
long long phi(long long m, int n)
{
if (n == 0) return m;
if (primes[n - 1] >= m) return 1;
if (m < MAXM && n < MAXN) return dp[n][m];
return phi(m, n - 1) - phi(m / primes[n - 1], n - 1);
}
long long Lehmer(long long m)
{
if (m < MAX) return counter[m];
long long w, res = 0;
int i, a, s, c, x, y;
s = sqrt(0.9 + m), y = c = cbrt(0.9 + m);
a = counter[y], res = phi(m, a) + a - 1;
for (i = a; primes[i] <= s; i++) res = res - Lehmer(m / primes[i]) + Lehmer(primes[i]) - 1;
return res;
}
}
long long solve(long long n)
{
int i, j, k, l;
long long x, y, res = 0;
for (i = 0; i < pcf::len; i++)
{
x = pcf::primes[i], y = n / x;
if ((x * x) > n) break;
res += (pcf::Lehmer(y) - pcf::Lehmer(x));
}
for (i = 0; i < pcf::len; i++)
{
x = pcf::primes[i];
if ((x * x * x) > n) break;
res++;
}
return res;
}
int main()
{
pcf::init();
long long n, res;
while (scanf("%lld", &n) != EOF)
{
//res = solve(n);
printf("%lld\n",pcf::Lehmer(n));
//printf("%lld\n", res);
}
return 0;
}
模板二代码
#include<cstdio>
#include<cstring>
#include<cctype>
#include<cmath>
#include<set>
#include<map>
#include<list>
#include<queue>
#include<deque>
#include<stack>
#include<string>
#include<vector>
#include<iostream>
#include<algorithm>
#include<stdlib.h>
#include<time.h>
using namespace std;
typedef long long LL;
const int INF=2e9+1e8;
const int MOD=1e9+7;
const int MAXSIZE=1e6+5;
const double eps=0.0000000001;
void fre()
{
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
}
#define memst(a,b) memset(a,b,sizeof(a))
#define fr(i,a,n) for(int i=a;i<n;i++)
LL f[320000],g[320000],n;
void init()
{
LL i,j,m;
for(m=1; m*m<=n; ++m) f[m]=n/m-1;
for(i=1; i<=m; ++i) g[i]=i-1;
for(i=2; i<=m; ++i)
{
if(g[i]==g[i-1])continue;
for(j=1; j<=min(m-1,n/i/i); ++j)
{
if(i*j<m)f[j]-=f[i*j]-g[i-1];
else f[j]-=g[n/i/j]-g[i-1];
}
for(j=m; j>=i*i; --j)g[j]-=g[j/i]-g[i-1];
}
}
int main()
{
while(scanf("%I64d",&n)&&n)
{
init();
cout<<f[1]<<endl;
}
return 0;
}
--> 点击打开链接题解