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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
a = xz space and space 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:
代码语言:javascript复制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[n]);
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。。。