Count primes
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; }
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