大家好,欢迎来到IT知识分享网。
1473: L先生与质数V3
Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 1348 Solved: 147
[Submit][Status][Web Board]
Description
在解决了上一个质数问题之后,L先生依然不甘心,他还想计算下更多范围内的质数,你能帮助他吗?
Input
有多组测试例。(测试例数量<70)
每个测试例一行,输入一个数字n(0<n<=3000000),输入0表示结束。
每个测试例一行,输入一个数字n(0<n<=3000000),输入0表示结束。
Output
输出测试例编号和第N个质数。
Case X: Y
Case X: Y
Sample Input
1
2
3
4
10
100
0
Sample Output
Case 1: 2
Case 2: 3
Case 3: 5
Case 4: 7
Case 5: 29
Case 6: 541
每次二分判断即可;
(参考别人的代码)
#include <algorithm> #include <iostream> #include <cstdio> #include <cmath> #include <cstring> typedef long long LL; const int N = 5e6 + 2;//通过知道前面的n^1/3的=质数可以推断后面n^2/3的质数所以可以适当减小 bool np[N]; int prime[N], pi[N]; int getprime() { int cnt = 0; np[0] = np[1] = true; pi[0] = pi[1] = 0; for(int i = 2; i < N; ++i) { if(!np[i]) prime[++cnt] = i; pi[i] = cnt; for(int j = 1; j <= cnt && i * prime[j] < N; ++j) { np[i * prime[j]] = true; if(i % prime[j] == 0) break; } } return cnt; } const int M = 7;//为了减小内存可以不过是质数 const int PM = 2 * 3 * 5 * 7 * 11 * 13 * 17;//为了减小内存可以不过要按质数减小如去掉17 int phi[PM + 1][M + 1], sz[M + 1]; void init() { getprime(); sz[0] = 1; for(int i = 0; i <= PM; ++i) phi[i][0] = i; for(int i = 1; i <= M; ++i) { sz[i] = prime[i] * sz[i - 1]; for(int j = 1; j <= PM; ++j) phi[j][i] = phi[j][i - 1] - phi[j / prime[i]][i - 1]; } } int sqrt2(LL x) { LL r = (LL)sqrt(x - 0.1); while(r * r <= x) ++r; return int(r - 1); } int sqrt3(LL x) { LL r = (LL)cbrt(x - 0.1); while(r * r * r <= x) ++r; return int(r - 1); } LL getphi(LL x, int s) { if(s == 0) return x; if(s <= M) return phi[x % sz[s]][s] + (x / sz[s]) * phi[sz[s]][s]; if(x <= prime[s]*prime[s]) return pi[x] - s + 1; if(x <= prime[s]*prime[s]*prime[s] && x < N) { int s2x = pi[sqrt2(x)]; LL ans = pi[x] - (s2x + s - 2) * (s2x - s + 1) / 2; for(int i = s + 1; i <= s2x; ++i) ans += pi[x / prime[i]]; return ans; } return getphi(x, s - 1) - getphi(x / prime[s], s - 1); } LL getpi(LL x) { if(x < N) return pi[x]; LL ans = getphi(x, pi[sqrt3(x)]) + pi[sqrt3(x)] - 1; for(int i = pi[sqrt3(x)] + 1, ed = pi[sqrt2(x)]; i <= ed; ++i) ans -= getpi(x / prime[i]) - i + 1; return ans; } LL lehmer_pi(LL x) { if(x < N) return pi[x]; int a = (int)lehmer_pi(sqrt2(sqrt2(x))); int b = (int)lehmer_pi(sqrt2(x)); int c = (int)lehmer_pi(sqrt3(x)); LL sum = getphi(x, a) +(LL)(b + a - 2) * (b - a + 1) / 2; for (int i = a + 1; i <= b; i++) { LL w = x / prime[i]; sum -= lehmer_pi(w); if (i > c) continue; LL lim = lehmer_pi(sqrt2(w)); for (int j = i; j <= lim; j++) sum -= lehmer_pi(w / prime[j]) - (j - 1); } return sum; } int main()//因为统计了质数个数可以用在线判断用二分 { LL n,l,r,mid; int sum=1; init(); while(scanf("%lld",&n)!=EOF) { if(n==0) break; l=1,r=50000000; while(l<r) { mid=(l+r)/2; LL t=lehmer_pi(mid); if(t>=n) r=mid; else l=mid+1; } printf("Case %d: %lld\n",sum++,l); } return 0; }
免责声明:本站所有文章内容,图片,视频等均是来源于用户投稿和互联网及文摘转载整编而成,不代表本站观点,不承担相关法律责任。其著作权各归其原作者或其出版社所有。如发现本站有涉嫌抄袭侵权/违法违规的内容,侵犯到您的权益,请在线联系站长,一经查实,本站将立刻删除。 本文来自网络,若有侵权,请联系删除,如若转载,请注明出处:https://yundeesoft.com/32263.html