hdu 2478||3090 欧拉函数
2015-09-21 15:17
267 查看
Farey Sequence
Description
The Farey Sequence Fn for any integer n with n >= 2 is the set of irreducible rational numbers a/b with 0 < a < b <= n and gcd(a,b) = 1 arranged in increasing order. The first few are
F2 = {1/2}
F3 = {1/3, 1/2, 2/3}
F4 = {1/4, 1/3, 1/2, 2/3, 3/4}
F5 = {1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5}
You task is to calculate the number of terms in the Farey sequence Fn.
Input
There are several test cases. Each test case has only one line, which contains a positive integer n (2 <= n <= 106). There are no blank lines between cases. A line with a single 0 terminates the input.
Output
For each test case, you should output one line, which contains N(n) ---- the number of terms in the Farey sequence Fn.
Sample Input
Sample Output
code:完整的注释部分是另一种求解
Visible Lattice Points
Description
A lattice point (x, y) in the first quadrant (x and y are integers greater than or equal to 0), other than the origin, is visible from the origin if the
line from (0, 0) to (x, y) does not pass through any other lattice point. For example, the point (4, 2) is not visible since the line from the origin passes through (2, 1). The figure below shows the points
(x, y) with 0 ≤ x, y ≤ 5 with lines from the origin to the visible points.
Write a program which, given a value for the size, N, computes the number of visible points (x, y) with 0 ≤ x, y ≤ N.
Input
The first line of input contains a single integer C (1 ≤ C ≤ 1000) which is the number of datasets that follow.
Each dataset consists of a single line of input containing a single integer N (1 ≤ N ≤ 1000), which is the size.
Output
For each dataset, there is to be one line of output consisting of: the dataset number starting at 1, a single space, the size, a single space and the number of visible points for that size.
Sample Input
Sample Output
思路 转 :先画一条(0, 0)到(n, n)的线,把图分成两部分,两部分是对称的,只需算一部分就好。
取右下半,这一半里的点(x, y)满足x >= y
可以通过欧拉函数计算第k列有多少点能够连到(0, 0)
若x与k的最大公约数d > 1,则(0, 0)与(x, k)点的连线必定会通过(x/d, k/d),就被挡住了
所以能连的线的数目就是比k小的、和k互质的数的个数,然后就是欧拉函数。
code:
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 0 | Accepted: 0 |
The Farey Sequence Fn for any integer n with n >= 2 is the set of irreducible rational numbers a/b with 0 < a < b <= n and gcd(a,b) = 1 arranged in increasing order. The first few are
F2 = {1/2}
F3 = {1/3, 1/2, 2/3}
F4 = {1/4, 1/3, 1/2, 2/3, 3/4}
F5 = {1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5}
You task is to calculate the number of terms in the Farey sequence Fn.
Input
There are several test cases. Each test case has only one line, which contains a positive integer n (2 <= n <= 106). There are no blank lines between cases. A line with a single 0 terminates the input.
Output
For each test case, you should output one line, which contains N(n) ---- the number of terms in the Farey sequence Fn.
Sample Input
2 3 4 5 0
Sample Output
1 3 5 9
code:完整的注释部分是另一种求解
#include <iostream> #include<string.h> #include<algorithm> #include<stdio.h> using namespace std; const int size=1000005; bool ok[size]; int ph[size]; int prime[size]; __int64 sum[size]; int pn; int n; //void getph() //{ // int i,j; // pn=0; // for(i=2;i<size;i++) // { // if(!ok[i]) // { // prime[pn++]=i; // ph[i]=i-1; // } // for(j=0;(j<pn)&&(i*prime[j]<size);j++) // { // ok[i*prime[j]]=1; // if((i%prime[j])==0) // { // ph[i*prime[j]]=ph[i]*prime[j]; // break; // } // else // ph[i*prime[j]]=ph[i]*(prime[j]-1); // // } // } //} void Euler() { int i,j; memset(ph,0,sizeof(ph)); for(i=2;i<=1000002;i++) { if(!ph[i]) { for(j=i;j<1000002;j+=i) { if(!ph[j]) ph[j]=j; ph[j]=(ph[j]/i)*(i-1); } } } } int main() { ph[1]=1; // getph(); Euler(); for(int i=2;i<size;i++) sum[i]=sum[i-1]+ph[i]; while(scanf("%d",&n)!=EOF&&(n!=0)) printf("%I64d\n",sum ); return 0; }
Visible Lattice Points
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 6113 | Accepted: 641 |
A lattice point (x, y) in the first quadrant (x and y are integers greater than or equal to 0), other than the origin, is visible from the origin if the
line from (0, 0) to (x, y) does not pass through any other lattice point. For example, the point (4, 2) is not visible since the line from the origin passes through (2, 1). The figure below shows the points
(x, y) with 0 ≤ x, y ≤ 5 with lines from the origin to the visible points.
Write a program which, given a value for the size, N, computes the number of visible points (x, y) with 0 ≤ x, y ≤ N.
Input
The first line of input contains a single integer C (1 ≤ C ≤ 1000) which is the number of datasets that follow.
Each dataset consists of a single line of input containing a single integer N (1 ≤ N ≤ 1000), which is the size.
Output
For each dataset, there is to be one line of output consisting of: the dataset number starting at 1, a single space, the size, a single space and the number of visible points for that size.
Sample Input
4 2 4 5 231
Sample Output
1 2 5 2 4 13 3 5 21 4 231 32549
思路 转 :先画一条(0, 0)到(n, n)的线,把图分成两部分,两部分是对称的,只需算一部分就好。
取右下半,这一半里的点(x, y)满足x >= y
可以通过欧拉函数计算第k列有多少点能够连到(0, 0)
若x与k的最大公约数d > 1,则(0, 0)与(x, k)点的连线必定会通过(x/d, k/d),就被挡住了
所以能连的线的数目就是比k小的、和k互质的数的个数,然后就是欧拉函数。
code:
#include<stdio.h> #include<math.h> int euler(int x) { int i, res=x,tmp; tmp=(int)sqrt(x * 1.0) + 1; for (i = 2; i <tmp; i++) if(x%i==0) { res = res / i * (i - 1); while (x % i == 0) x /= i; } if (x > 1) res = res / x * (x - 1); return res; } int main() { int test,n,i,ans,_case=1; scanf("%d",&test); while(test--) { ans=0; scanf("%d",&n); for(i=2;i<=n;i++)//cong 2 kaishi ans+=euler(i); printf("%d %d %d\n",_case++,n,ans*2+3);//di yi lie you 2 ge ,zhongjian you yige } return 0; }
相关文章推荐
- 近年来作品整理——杂
- crash率考核的几个考虑因素
- android 程序外启动其他应用
- MUI - myStorage在ios safari无痕浏览模式下的解决方案
- HDU 4044 GeoDefense (树形DP,混合经典)
- 职场准备(一)
- UVA 10006 Carmichael Numbers (暴力+快速幂取模)
- JS有参函数的创建和调用
- centos6.5安装jetty服务器
- LINK - Azure - Session/Cache State Management
- ajaxfileupload 附加参数
- 将String转换成InputStream
- Linux 基础知识:变量键盘读取、数组与声明
- 定义接口类型的引用变量有什么好处?
- 快速实现 ListView下拉,图片放大刷新操作
- 【bzoj2751】【HAOI2012】【容易题】【数学】
- iOS9 兼容那点事
- linux/UNIX上的inode、硬链接以及软连接
- ios开发者修养
- OCP-V13-686