一次不严谨的C++、C、Pascal、Rust等对比

起因

现在ACM用得多的基本上就两种语言，C++和Python3，还有部分Java，但是当年ACM必学的Pascal、新近流行的rust也有人用，只不过用户很少。

就以一道codeforce上的算法小题为样本，来对比一样用户数量、执行效率、易写程度。

题目

https://codeforces.com/contest/1997/problem/C

样本总量：20150

用户数量对比

C++20：7800

C++17：9150

C++11：1550

Python3：1050

Python2：1

Haskell：2

C：45

C#：17

Go：14

Java21：450

Java8：180

Kotlin：9

Pascal(FPC)：3

Rust：30

各语言数量差异很大，很直观。基本上是C++统治的时代了。

语言代码量对比

就取各语言中代码量最少的一个

C++20

cpp 复制代码

#include<bits/stdc++.h>
using namespace std;
int t,n,ans;
string st;
int main(){
	cin>>t;
	while(t--){
		cin>>n>>st;
		ans=n/2;
		for(int i=0;i<n;i++)
			if(st[i]=='(')
				ans+=2;
		cout<<ans<<endl;
	}
	return 0;
}

C

cpp 复制代码

#include <stdio.h>
 
char s[200100];
int main(void){
	int tt, ii;
	int ans;
	int n, i;
	
	scanf("%d", &tt);
	for (ii=1; ii<=tt; ii++){
		scanf("%d", &n);
		ans=n/2;
		scanf("%s", s+1);
		for (i=1; i<=n; i++){
			if (s[i]=='('){
				ans+=2;	
			}
		}
		printf("%d\n", ans);
	}
	
    return 0;
}

C++14

cpp 复制代码

#include<bits/stdc++.h>
using namespace std;
int main(){
	string st;
	int t,n,ans;
	cin>>t;
	while(t--){
		cin>>n>>st;
		ans=n/2;
		for(int i=0;i<n;i++)
			if(st[i]=='(')
				ans+=2;
		cout<<ans<<endl;
	}
	return 0;
}

Python3

python 复制代码

for _ in[0]*int(input()):print((y:=int(input()))//2+2*input().count('('))

Pascal

Delphi 复制代码

Var
  t,i,n,dem1:Longint;
  kq:Int64;
  a:Array[0..300000] of Longint;
Procedure giai;
Var i:Longint;
    s:ansistring;
Begin
  Readln(n);
  Readln(s);
  {d1:=0;d2:=0;
  For i:=1 to length(s) do if i mod 2=0 then
    If s[i]='(' then inc(d1) else inc(d2); Inc(d1);}
  dem1:=1;
  a[dem1]:=1;
  kq:=0;
  For i:=2 to length(s) do
    If s[i]=')' then
      Begin
        kq:=kq+i-a[dem1];
        Dec(dem1);
      End
    Else if s[i]='(' then
      Begin
        Inc(dem1);
        a[dem1]:=i;
      End
    Else if dem1>0 then
      Begin
        kq:=kq+i-a[dem1];
        Dec(dem1);
      End
    Else
      Begin
        Inc(dem1);
        a[dem1]:=i;
      End;
  Writeln(kq);
End;
BEGIN
 // Assign(input,'contest168c.inp');Reset(input);
 // Assign(output,'contest168c.out');Rewrite(output);
  Readln(t);
  For i:=1 to t do giai();
END.

Rust

rust 复制代码

use std::io::*;

fn main() {
    let s = read_to_string(stdin()).unwrap();
    let mut it = s.lines();
    let mut nx = || it.next().unwrap();

    let t: usize = nx().parse().unwrap();
    for _ in 0..t {
        let _: usize = nx().parse().unwrap();
        let s = nx().chars();
        let mut sum = 0;
        let mut st = vec![];
        for (i,ch) in s.enumerate() {
            if ch == '_' {
                if st.is_empty() {
                    st.push(i);
                } else {
                    sum += i - st.pop().unwrap();
                }
            } else if ch == '(' {
                st.push(i);
            } else {
                sum += i - st.pop().unwrap();
            }
        }
        println!("{sum}");
    }
}

Java

java 复制代码

import java.util.*;

public class RanitMukherjee {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while(t-- > 0) {
            int n = sc.nextInt();
            char[] arr = sc.next().toCharArray();
            int count = 0;
            for(int i = 1; i < n; i+=2) {
                if(arr[i] == '(') count += 3;
                else count++;
            }
            System.out.println(count);
        }
        sc.close();
    }
}

Haskell

Haskell 复制代码

import qualified Data.ByteString.Char8 as C
import Data.Maybe

constOpen :: Char
constOpen = '('

constClose :: Char
constClose = ')'

constBlank :: Char
constBlank = '_'

parseInt :: C.ByteString -> Int
parseInt x = fst (fromJust (C.readInt x))

getInts :: IO [Int]
getInts = do
	line <- C.getLine
	let res = map (parseInt) (C.words line)
	return res

parseStr :: C.ByteString -> String
parseStr x = C.unpack x

getStrs :: IO [String]
getStrs = do
	line <- C.getLine
	let res = map (parseStr) (C.words line)
	return res

fill :: String -> String
fill [x] = [x]
fill (x : y : rest)
	| (y == constBlank && x == constOpen) = x : constClose : fill rest
	| (y == constBlank && x == constClose) = x : constOpen : fill rest
	| otherwise = fill (y : rest)

getBracketPairs :: [(Char, Int)] -> [Int] -> [(Int, Int)]
getBracketPairs [] _ = []
getBracketPairs ((c, i) : rest) stack
	| (c == constOpen) = getBracketPairs rest (i : stack)
	| otherwise = (i, head stack) : getBracketPairs rest (tail stack)

ans :: IO ()
ans = do
	[n] <- getInts
	[s] <- getStrs
	let sFill = constOpen : fill (tail s)
	let pairs = zip sFill [1 .. ]
	let bracketPairs = getBracketPairs pairs []
	let dists = map (\(x, y) -> x - y) bracketPairs
	let res = sum dists
	print res
	return ()

main :: IO ()
main = do
	[t] <- getInts
	sequence (replicate t ans)
	return ()

Kotlin

Kotlin 复制代码

import java.util.LinkedList
 
fun main() {
    repeat(readln().toInt()) {
        val n = readln().toInt()
        val s = readln()
 
        var ans = 0L
        val bracketPositions = LinkedList<Int>()
        for (i in s.indices) {
            var c = s[i]
            if (c == '_') {
                c = if (bracketPositions.isEmpty()) '(' else ')'
            }
            if (c == ')') {
                ans += i - bracketPositions.pollLast()
            }
            else
                bracketPositions.addLast(i)
        }
        println(ans)
    }
}

C#

cs 复制代码

using System.Text;

namespace EvenPositions;
class Program
{
    private static readonly StreamReader reader = new StreamReader(Console.OpenStandardInput(1024 * 10), Encoding.ASCII, false, 1024 * 10);
    private static readonly StreamWriter writer = new StreamWriter(Console.OpenStandardOutput(1024 * 10), Encoding.ASCII, 1024 * 10);
    private static void Main(string[] args)
    {
        int t = int.Parse(reader.ReadLine());
        for (int i = 0; i < t; i++)
        {
            writer.WriteLine(Solve());
        }
        writer.Flush();
    }

    private static int Solve()
    {
        reader.ReadLine();
        string s = reader.ReadLine();
        int ans = 0;
        for (int i = 1; i < s.Length; i += 2)
        {
            if (s[i] == ')') ans += 1;
            else ans += 3;
        }
        return ans;
    }
}

总结

一个语言用户数量的多少，最相关的还是难易程度，同样的代码写起来越简单肯定越受人喜欢；其次就是运行效率，尽管简单，过不了运行测试也不行；再就是用户基数，用的人多，学习资料才多，而用户基数就是前面两点确定的。