The document discusses standard input and output in programming. It explains that standard input comes from keyboard input while standard output displays on the screen. It then covers different methods for file input/output like fopen, fstream, freopen, and their performance. The document concludes with recommendations to use buffered reading/writing techniques like fgets/puts for better performance compared to scanf/printf.

1. About I/O
郭至軒（KuoE0）
KuoE0.tw@gmail.com
KuoE0.ch

2. Attribution-ShareAlike 3.0 Unported
(CC BY-SA 3.0)
http://creativecommons.org/licenses/by-sa/3.0/
Latest update: Feb 27, 2013

3. Standard Input & Output
- 標準輸入
- 標準輸出

4. Standard Input & Output
- 標準輸入
- 標準輸出

5. Standard Input & Output
- 標準輸入
- 標準輸出

6. 標準輸入→由 盤輸入
標準輸出→由螢幕輸出

7. input file output file

8. Multiple Test Case
12 3
34 7
56 11
78 15
input file output file

9. Continuous Processing
output
calculate
result
read one
start no data
test case
end

10. start read: 1 2 calculate write: 3
read: 5 6 write: 7 calculate read: 3 4
calculate write: 11 read: 7 8 calculate
end write: 15

11. End Of File
若題目未指定測資終止條件，則為判斷 EOF 作為終止
條件！
scanf fgets cin
while (scanf() != EOF) while (fgets() != 0) while (cin >> x)
{ { {
... ... ...
} } }

12. File Input & Output
fopen fstream (C++ only)
ifstream in =
FILE *in = fopen(“inputfile”); ifstream(“inputfile”);
FILE *out =
fopen(“outputfile”); ofstream out =
ofstream(“outputfile”);
fscanf(in, ...);
fprintf(out, ...); in >> x;
out << x;
fclose(in);
fclose(out); in.close();
out.close();

13. If you write the code like
previous page in contest, you
will lose the contest.

14. freopen
redirect the file I/O to standard I/O
freopen(“inputfile”, “r”, stdin);
freopen(“output”, “w”, stdout);
scanf(...);
printf(...);

15. I/O Performance

16. scanf & cin
0
10^4
0
0.01
10^5
0.04
0.14
10^6
0.41
1.27
10^7
3.45
0 sec 1 sec 2 sec 3 sec 4 sec
scanf cin

17. printf & cout
0
10^4
0.01
0.01
10^5
0.04
0.11
10^6
0.42
1.08
10^7
4.02
0 sec 1.25 sec 2.5 sec 3.75 sec 5 sec
printf cout

18. Improve Stream I/O
Stream I/O need to keep
themselves in sync with
the underlying C library.
Add this line in code,
std::ios::sync_with_stdio(false);
if you won’t use C I/O.

19. cin without sync
0
10^4 0
0
0.01
10^5 0.04
0.02
0.14
10^6 0.41
0.19
1.27
10^7 3.45
1.70
0 sec 1 sec 2 sec 3 sec 4 sec
scanf cin
cin without sync

20. cout without sync
0
10^4 0.01
0.01
0.01
10^5 0.04
0.04
0.11
10^6 0.42
0.36
1.08
10^7 4.02
3.49
0 sec 1.25 sec 2.5 sec 3.75 sec 5 sec
printf cout
cout without sync

21. Stream I/O is Still Slow
Recommend
to use C I/O.

22. Buffered Technique
Base on fgets/puts function.
fgets/puts are faster than scanf/printf.

23. Buffered Read
Read mass data, and parse by self.
char buf[ 1000 ];
int a, b;
fgets( buf, sizeof( buf ), stdin );
sscanf( buf, “%d %d”, &a, &b );

24. Buffered Write
Store mass data into temporal buffer,
and write them once.
char buf[ 1000 ];
int ret = a + b;
sprintf( buf, “%d”, ret );
puts( buf );

25. Buffering Technique
0
0
10^4 0
0
0.01
0.04
10^5 0.02
0.01
0.14
0.41
10^6 0.19
0.05
1.27
3.45
10^7 1.70
0.53
0 sec 1 sec 2 sec 3 sec 4 sec
scanf cin
cin without sync fgets

26. Buffering Technique
0
0.01
10^4 0.01
0
0.01
0.04
10^5 0.04
0.02
0.11
0.42
10^6 0.36
0.16
1.08
4.02
10^7 3.49
1.71
0 sec 1.25 sec 2.5 sec 3.75 sec 5 sec
printf cout
cout without sync puts

27. Advanced Parsing Skill
strtok Split string into tokens.
char* strtok( char *str, const char *delimiters );
str: 欲切割之字串
delimiters: 分隔字符字串
return value: 指向當前切割字串之指
標，若切割完畢則回傳 NULL

28. strtok
original string:
A “corpus” is a collection of texts of written (or
spoken) language presented in electronic form.
take out all words:
A corpus is a
collection of texts of
written or spoken language
presented in electronic form

29. char str = “A “corpus” is a collection of texts of
written (or spoken) language presented in electronic
form.”
for ( char *token = strtok( str, “ ”().” ); token !=
NULL; token = strtok( NULL, “ ”().” ) ) {
puts( token );
}

30. start position
delimiters: “ ”().”
global pointer
A “ c o r p u s ” i s a c o l l
e c t i o n o f t e x t s o f w
r e t t e n ( o r s p o k e n ) l
a n g u a g e p r e s e n t e d i n
e l e t r o n i c f o r m . 0

31. start position
delimiters: “ ”().”
global pointer
A 0 “ c o r p u s ” i s a c o l l
e c t i o n o f t e x t s o f w
r e t t e n ( o r s p o k e n ) l
a n g u a g e p r e s e n t e d i n
e l e t r o n i c f o r m . 0

32. Return address of start position.
“A”

33. start position
delimiters: “ ”().”
global pointer
A 0 “ c o r p u s ” i s a c o l l
e c t i o n o f t e x t s o f w
r e t t e n ( o r s p o k e n ) l
a n g u a g e p r e s e n t e d i n
e l e t r o n i c f o r m . 0

34. start position
delimiters: “ ”().”
global pointer
A 0 0 c o r p u s 0
“ ” i s a c o l l
e c t i o n o f t e x t s o f w
r e t t e n ( o r s p o k e n ) l
a n g u a g e p r e s e n t e d i n
e l e t r o n i c f o r m . 0

35. Return address of start position.
“corpus”

36. start position
delimiters: “ ”().”
global pointer
A 0 0 c o r p u s 0 i s a c o l l
e c t i o n o f t e x t s o f w
r e t t e n ( o r s p o k e n ) l
a n g u a g e p r e s e n t e d i n
e l e t r o n i c f o r m . 0

37. start position
delimiters: “ ”().”
global pointer
A 0 0 c o r p u s 0 0 i s 0 a c o l l
e c t i o n o f t e x t s o f w
r e t t e n ( o r s p o k e n ) l
a n g u a g e p r e s e n t e d i n
e l e t r o n i c f o r m . 0

38. Return address of start position.
“is”

39. start position
delimiters: “ ”().”
global pointer
A 0 0 c o r p u s 0 0 i s 0 a c o l l
e c t i o n o f t e x t s o f w
r e t t e n ( o r s p o k e n ) l
a n g u a g e p r e s e n t e d i n
e l e t r o n i c f o r m . 0

40. start position
delimiters: “ ”().”
global pointer
A 0 0 c o r p u s 0 0 i s 0 a 0 c o l l
e c t i o n o f t e x t s o f w
r e t t e n ( o r s p o k e n ) l
a n g u a g e p r e s e n t e d i n
e l e t r o n i c f o r m . 0

41. Return address of start position.
“a”

42. After 5 minutes...

43. start position
delimiters: “ ”().”
global pointer
A 0 0 c o r p u s 0 0 i
“ ” s 0 a 0 c o l l
e c t i o n 0 o f 0 t e x t s 0 o f 0 w
r e t t e n 0 0 o r 0 s p o k e n 0 0 l
a n g u a g e 0 p r e s e n t e d 0 i n
0 e l e t r o n i c 0 f o r m . 0

44. start position
delimiters: “ ”().”
global pointer
A 0 0 c o r p u s 0 0 i
“ ” s 0 a 0 c o l l
e c t i o n 0 o f 0 t e x t s 0 o f 0 w
r e t t e n 0 0 o r 0 s p o k e n 0 0 l
a n g u a g e 0 p r e s e n t e d 0 i n
0 e l e t r o n i c 0 f o r m 0 0
.

45. Return address of start position.
“form”

46. start position
delimiters: “ ”().”
global pointer
A 0 0 c o r p u s 0 0 i
“ ” s 0 a 0 c o l l
e c t i o n 0 o f 0 t e x t s 0 o f 0 w
r e t t e n 0 0 o r 0 s p o k e n 0 0 l
a n g u a g e 0 p r e s e n t e d 0 i n
0 e l e t r o n i c 0 f o r m 0 0
.

47. Return address of start position.
NULL

48. • include <cstring> or <string.h>
• first time use string variable as
parameter to setup global pointer
• others use NULL as parameter to avoid
changing global pointer
• strtok will modify the original string
• whitespace character (n, r, t ...)

49. Practice Now
POJ 3981 - 字符串替换

50. Thank You for Your
Listening.