计算阶乘中的位数-输入/输出性能问题 [英] Count number of digits in factorial - Input/Ouput performance issue
问题描述
我正在spoj平台上解决任务-计算阶乘中的位数.我找到了Kamenetsky公式并实现了它:
I am solving task on spoj platform - count number of digits in factorial. I found Kamenetsky Formula and implemented it:
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,我使用注释的代码(流),因为我认为它比实际的代码(没有注释)要慢,所以我进行了更改,但仍然超出了时间限制.我怎样才能让它更快?
first I used commented code (streams) and as I thought it's slower than actual one (without comments) so I changed but still getting exceed time limit. How can I make it faster?
示例输入为(第一行是测试数量):
Example input is (first line is number of tests):
3
1
10
100
和预期输出:
1
7
158
推荐答案
我认为由于I/O缓慢而超出了时间限制.这主要是因为System.out.println的基础PrintStream.在这篇文章中找到更多详细信息 why-is-system-out-println-so-慢.您可以参考下面的快速I/O模板,这将有助于解决此问题.
I think the time limit is exceeding because of the slow I/O. This is mainly because of System.out.println's underlying PrintStream. Find more details in this post why-is-system-out-println-so-slow. You can refer to the below Fast I/O template which will help solve this problem.
参考- Java中的快速I/O
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);
}
}
}
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