计算阶乘的位数 - Input/Ouput 性能问题
Count number of digits in factorial - Input/Ouput performance issue
我正在 spoj 平台上解决任务 - 计算阶乘的位数。
我找到了 Kamenetsky 公式并实现了它:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
int numOfTests = Integer.parseInt(in.readLine());
for(int i = 0; i < numOfTests; i++) {
System.out.println(KamenetskyFormula(Integer.parseInt(in.readLine())));
}
/*
in.lines()
.limit(numOfTests)
.map(n -> Integer.parseInt(n))
.forEach(n -> System.out.println(KamenetskyFormula(n)));*/
}
private static long KamenetskyFormula(int n) {
if (n < 2) {
return 1;
}
double x = n * Math.log10(n / Math.E) + Math.log10(2 * Math.PI * n) / 2.0;
return (long) (Math.floor(x) + 1);
}
}
首先我使用了注释代码(流),因为我认为它比实际代码(没有注释)慢所以我改变了但仍然超过时间限制。我怎样才能让它更快?
示例输入是(第一行是测试数):
3
1
10
100
和预期输出:
1
7
158
我认为由于速度慢 I/O 超过了时间限制。这主要是因为System.out.println的底层PrintStream。在此 post why-is-system-out-println-so-slow 中查找更多详细信息。您可以参考下面的快速I/O模板,这将有助于解决这个问题。
参考 - Fast I/O in java
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
public class YourClassName {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
TaskA solver = new TaskA();
int testCount = Integer.parseInt(in.next());
for (int i = 1; i <= testCount; i++)
solver.solve(i, in, out);
out.close();
}
static class TaskA {
public void solve(int testNumber, InputReader sc, PrintWriter w) {
/*your logic goes here
use w.println here to print the output which is usually faster
*/
}
}
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private InputReader.SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public int nextInt() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String readString() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public String next() {
return readString();
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
}
我正在 spoj 平台上解决任务 - 计算阶乘的位数。 我找到了 Kamenetsky 公式并实现了它:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main {
public static void main(String[] args) throws IOException {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
int numOfTests = Integer.parseInt(in.readLine());
for(int i = 0; i < numOfTests; i++) {
System.out.println(KamenetskyFormula(Integer.parseInt(in.readLine())));
}
/*
in.lines()
.limit(numOfTests)
.map(n -> Integer.parseInt(n))
.forEach(n -> System.out.println(KamenetskyFormula(n)));*/
}
private static long KamenetskyFormula(int n) {
if (n < 2) {
return 1;
}
double x = n * Math.log10(n / Math.E) + Math.log10(2 * Math.PI * n) / 2.0;
return (long) (Math.floor(x) + 1);
}
}
首先我使用了注释代码(流),因为我认为它比实际代码(没有注释)慢所以我改变了但仍然超过时间限制。我怎样才能让它更快?
示例输入是(第一行是测试数):
3
1
10
100
和预期输出:
1
7
158
我认为由于速度慢 I/O 超过了时间限制。这主要是因为System.out.println的底层PrintStream。在此 post why-is-system-out-println-so-slow 中查找更多详细信息。您可以参考下面的快速I/O模板,这将有助于解决这个问题。
参考 - Fast I/O in java
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
public class YourClassName {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
TaskA solver = new TaskA();
int testCount = Integer.parseInt(in.next());
for (int i = 1; i <= testCount; i++)
solver.solve(i, in, out);
out.close();
}
static class TaskA {
public void solve(int testNumber, InputReader sc, PrintWriter w) {
/*your logic goes here
use w.println here to print the output which is usually faster
*/
}
}
static class InputReader {
private InputStream stream;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
private InputReader.SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public int nextInt() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public String readString() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
public String next() {
return readString();
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
}