您的位置:首页 > 其它

挑战程序竞赛系列(44):4.1计数 欧拉函数

2017-09-01 08:52 423 查看

挑战程序竞赛系列(44):4.1计数 欧拉函数

详细代码可以fork下Github上leetcode项目,不定期更新。

练习题如下:

POJ 2407: Relatives

POJ 2407: Relatives

条件解析:

a=xz and b=xy,不存在x>1,z>0,y>0

说白了,两两互素,比如 a = 4, b = 9

代码如下:

import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Main{

String INPUT = "./data/judge/201709/P2407.txt";

public static void main(String[] args) throws IOException {
new Main().run();
}

static final int MAX_N = 10000 + 16;
int[] euler = new int[MAX_N];

public void euler_phi(){
for (int i = 0; i < MAX_N; ++i) euler[i] = i;
for (int i = 2; i < MAX_N; ++i){
if (euler[i] == i){
for (int j = i; j < MAX_N; j += i){
euler[j] = euler[j] / i * (i - 1);
}
}
}
}

public int phi(int n){
int res = n;
for (int i = 2; i < n / i; ++i){
if (n % i == 0){
res = res / i * (i - 1);
for (; n % i == 0; n /= i);
}
}
if (n != 1){
res = res / n * (n - 1);
}
return res;
}

void solve() {

euler_phi();

while (true){
int n = ni();
if (n == 0) break;
if (n < MAX_N)
out.println(euler
);
else
out.println(phi(n));
}
}

FastScanner in;
PrintWriter out;

void run() throws IOException {
boolean oj;
try {
oj = !System.getProperty("user.dir").equals("F:\\java_workspace\\leetcode");
} catch (Exception e) {
oj = System.getProperty("ONLINE_JUDGE") != null;
}

InputStream is = oj ? System.in : new FileInputStream(new File(INPUT));
in = new FastScanner(is);
out = new PrintWriter(System.out);
long s = System.currentTimeMillis();
solve();
out.flush();
if (!oj) {
System.out.println("[" + (System.currentTimeMillis() - s) + "ms]");
}
}

public boolean more() {
return in.hasNext();
}

public int ni() {
return in.nextInt();
}

public long nl() {
return in.nextLong();
}

public double nd() {
return in.nextDouble();
}

public String ns() {
return in.nextString();
}

public char nc() {
return in.nextChar();
}

class FastScanner {
BufferedReader br;
StringTokenizer st;
boolean hasNext;

public FastScanner(InputStream is) throws IOException {
br = new BufferedReader(new InputStreamReader(is));
hasNext = true;
}

public String nextToken() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (Exception e) {
hasNext = false;
return "##";
}
}
return st.nextToken();
}

String next = null;

public boolean hasNext() {
next = nextToken();
return hasNext;
}

public int nextInt() {
if (next == null) {
hasNext();
}
String more = next;
next = null;
return Integer.parseInt(more);
}

public long nextLong() {
if (next == null) {
hasNext();
}
String more = next;
next = null;
return Long.parseLong(more);
}

public double nextDouble() {
if (next == null) {
hasNext();
}
String more = next;
next = null;
return Double.parseDouble(more);
}

public String nextString() {
if (next == null) {
hasNext();
}
String more = next;
next = null;
return more;
}

public char nextChar() {
if (next == null) {
hasNext();
}
String more = next;
next = null;
return more.charAt(0);
}
}
}




啧啧,睡了一觉回来,跑不快了。。。
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: