C 和 Python简易性能对比

tang818875 · 发表于 2021-10-15 00:50:05

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C 和 Python的性能差别真TMD大，以求百万以内的素数为例，算法越低效效果越明显：

运行环境：相同

各版本的算法：

version 1: 逐个求余，最低效
version 2: 前一半求余，次之
version 3: 平方根求余, 最高效

结果：

version : lang: time: count
1:    c : 190s : 78499
2:    c:    98s: 78500
3:    c : 4s : 78667

1: python: 不试了，时间太长，等不了
2: python: 2190s: 78500
3: python: 10s: 78704

上代码：
C version1

#include <stdio.h>
#include <time.h>
int main() {
int num;
_Bool is_prime;
int start, end, count = 0;
start = time(NULL);
printf("Prime evaluation system ###\n");
printf("Please enter a legal whole number: ");
scanf("%d", &num);
printf("The prime between 0 and %d are: ", num);
for(int i=2; i<=num; i++) {
is_prime = 1;
for(int j=2; j<i; j++) {
if(!(i % j)) {
is_prime = 0;
break;
}
}
if(is_prime == 1) {
count ++;
printf("%d, ", i);
}
}
printf("\b\b \n");
printf("Version 2.0: The evaluating took %ds to finish and there\'re %d prime there.\n", (end = time(NULL)) - start, count);
return 0;
}

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C version 2

#include <stdio.h>
#include <time.h>
int main() {
int num;
_Bool is_prime;
int start, end, count = 0;
start = time(NULL);
printf("Prime evaluation system ###\n");
printf("Please enter a legal whole number: ");
scanf("%d", &num);
printf("The prime between 0 and %d are: ", num);
for(int i=1; i<=num; i++) {
is_prime = 1;
for(int j=2; j<i/2; j++) {
if(!(i % j)) {
is_prime = 0;
break;
}
}
if(is_prime == 1) {
count ++;
printf("%d, ", i);
}
}
printf("\b\b \n");
printf("The evaluating took %ds to finish and there\'re %d prime there.\n", (end = time(NULL)) - start, count);
return 0;
}

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C version 3

#include <stdio.h>
#include <time.h>
#include <math.h>
int main() {
int num;
_Bool is_prime;
int start, end, count = 0;
start = time(NULL);
printf("Prime evaluation system ###\n");
printf("Please enter a legal whole number: ");
scanf("%d", &num);
printf("The prime between 0 and %d are: ", num);
for(int i=1; i<=num; i++) {
is_prime = 1;
for(int j=2; j<sqrt(i); j++) {
if(!(i % j)) {
is_prime = 0;
break;
}
}
if(is_prime == 1) {
count ++;
printf("%d, ", i);
}
}
printf("\b\b \n");
printf("The evaluating took %ds to finish and there\'re %d prime there.\n", (end = time(NULL)) - start, count);
return 0;
}

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Python version 1:

import time
start = time.time()
print("Prime number evaluation system ###")
num = int(input("Please enter a legal numbers: "))
count = 0
print("The prime numbers between 0 and %d are: " % num, end="")
for i in range(2, num + 1):
is_prime = True
for j in range(2, i):
if not i % j:
is_prime = False
break
if is_prime == True:
count += 1
print("%d, " % i, end="")
print("\b\b ")
print("Version 1.0: The prime numbers evaluation tooks %ds to finish and there\'re %d primes there" % ((time.time() - start), count))

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Python version 2

import time
start = time.time()
print("Prime number evaluation system ###")
num = int(input("Please enter a legal numbers: "))
count = 0
print("The prime numbers between 0 and %d are: " % num, end="")
for i in range(2, num + 1):
is_prime = True
for j in range(2, i // 2):
if not i % j:
is_prime = False
break
if is_prime == True:
count += 1
print("%d, " % i, end="")
print("\b\b ")
print("Version 2.0: The prime numbers evaluation tooks %ds to finish and there\'re %d primes there" % ((time.time() - start), count))

复制代码

Python version 3

import time
import math
start = time.time()
print("Prime number evaluation system ###")
num = int(input("Please enter a legal numbers: "))
count = 0
print("The prime numbers between 0 and %d are: " % num, end="")
for i in range(2, num + 1):
is_prime = True
for j in range(2, int(math.sqrt(i))):
if not i % j:
is_prime = False
break
if is_prime == True:
count += 1
print("%d, " % i, end="")
print("\b\b ")
print("Version 2.0: The prime numbers evaluation tooks %ds to finish and there\'re %d primes there" % ((time.time() - start), count))

复制代码

当然啦，语言得看应用场景，脱离场景谈性能，有时无疑是追求大炮大小鸟---费劲

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