[ACM-ICPC] About I/O

Transcript

1. About I/O ֲࢸݢʢKuoE0ʣ 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 - ඪ४༌ೖ - ඪ४༌ग़
4. Standard Input & Output - ඪ४༌ೖ - ඪ४༌ग़
5. Standard Input & Output - ඪ४༌ೖ - ඪ४༌ग़
6. ඪ४༌ೖ→༝伴൫༌ೖ ඪ४༌ग़→༝ᦊນ༌ग़
7. input file output file
8. Multiple Test Case 1 2 3 4 5 6 7 8 3 7 11 15 input file output file
9. Continuous Processing read one test case no data start end output result calculate
10. read: 1 2 start calculate write: 3 read: 3 4 calculate write: 7 read: 5 6 calculate write: 11 read: 7 8 calculate write: 15 end
11. End Of File ए୊໨ະࢦఆଌࢿऴࢭᑍ݅ɼଇҝ൑Ꮧ EOF ࡞ҝऴࢭ ᑍ݅ʂ scanf fgets cin while (scanf() != EOF) { ... } while (fgets() != 0) { ... } while (cin >> x) { ... }
12. File Input & Output fopen fstream (C++ only) FILE *in = fopen(“inputfile”); FILE *out = fopen(“outputfile”); fscanf(in, ...); fprintf(out, ...); fclose(in); fclose(out); ifstream in = ifstream(“inputfile”); ofstream out = ofstream(“outputfile”); in >> x; out << x; 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 10^4 10^5 10^6 10^7 0 sec 1 sec 2 sec 3 sec 4 sec 3.45 0.41 0.04 0 1.27 0.14 0.01 0 scanf cin
17. printf & cout 10^4 10^5 10^6 10^7 0 sec 1.25 sec 2.5 sec 3.75 sec 5 sec 4.02 0.42 0.04 0.01 1.08 0.11 0.01 0 printf cout
18. Improve Stream I/O Stream I/O need to keep themselves in sync with the underlying C library. std::ios::sync_with_stdio(false); if you won’t use C I/O. Add this line in code,
19. cin without sync 10^4 10^5 10^6 10^7 0 sec 1 sec 2 sec 3 sec 4 sec 1.70 0.19 0.02 0 3.45 0.41 0.04 0 1.27 0.14 0.01 0 scanf cin cin without sync
20. cout without sync 10^4 10^5 10^6 10^7 0 sec 1.25 sec 2.5 sec 3.75 sec 5 sec 3.49 0.36 0.04 0.01 4.02 0.42 0.04 0.01 1.08 0.11 0.01 0 printf cout cout without sync
21. Stream I/O is Still Slow Recommend to use C I/O.
22. Buffered Technique fgets/puts are faster than scanf/printf. Base on fgets/puts function.
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 char buf[ 1000 ]; int ret = a + b; sprintf( buf, “%d”, ret ); puts( buf ); Store mass data into temporal buffer, and write them once.
25. Buffering Technique 10^4 10^5 10^6 10^7 0 sec 1 sec 2 sec 3 sec 4 sec 0.53 0.05 0.01 0 1.70 0.19 0.02 0 3.45 0.41 0.04 0 1.27 0.14 0.01 0 scanf cin cin without sync fgets
26. Buffering Technique 10^4 10^5 10^6 10^7 0 sec 1.25 sec 2.5 sec 3.75 sec 5 sec 1.71 0.16 0.02 0 3.49 0.36 0.04 0.01 4.02 0.42 0.04 0.01 1.08 0.11 0.01 0 printf cout cout without sync puts
27. Advanced Parsing Skill char* strtok( char *str, const char *delimiters ); strtok Split string into tokens. str: ཉ੾ׂ೭ࣈ۲ delimiters: ෼ִࣈූࣈ۲ return value: ࢦ޲ᙛલ੾ׂࣈ۲೭ࢦ ඪɼए੾ׂ׬ඟଇճၚ NULL
28. strtok A “corpus” is a collection of texts of written (or spoken) language presented in electronic form. original string: 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. 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 delimiters: “ \”().” global pointer start position
31. 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 \0 delimiters: “ \”().” global pointer start position
32. Return address of start position. “A”
33. 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 delimiters: “ \”().” global pointer start position
34. 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 \0 \0 delimiters: “ \”().” global pointer start position
35. Return address of start position. “corpus”
36. 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 delimiters: “ \”().” global pointer start position
37. 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 \0 \0 delimiters: “ \”().” global pointer start position
38. Return address of start position. “is”
39. 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 delimiters: “ \”().” global pointer start position
40. 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 \0 delimiters: “ \”().” global pointer start position
41. Return address of start position. “a”
42. After 5 minutes...
43. delimiters: “ \”().” A \0 \“ c o r p u s \” i s a 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 global pointer start position \0 \0 \0 \0 \0
44. delimiters: “ \”().” A \0 \“ c o r p u s \” i s a 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 global pointer start position \0 \0 \0 \0 \0 \0
45. Return address of start position. “form”
46. delimiters: “ \”().” A \0 \“ c o r p u s \” i s a 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 global pointer start position \0 \0 \0 \0 \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. POJ 3981 - ࣈූ۲ସ换 Practice Now
50. Thank You for Your Listening.