POJ1002: The first step in large data processing is I/O

This problem is not difficult, but http://www.aliyun.com/zixun/aggregation/20522.html "> Test data (http://www.ntnu.edu.tw/acm/ProblemSetArchive /b_us_eastcen/1999/index.html) Very savage, there are several 1 million lines of test data.

Not noticing the 2000 millisecond limit at first, the results of the first version of the program were written in many functions, avoiding global variables as much as possible, and using dynamic memory allocations and struct. Later found that always timed out. Check the other people's answers online, found that you have to do a stronger input format to assume, and then do not control too many style principles, as far as possible in a main to complete. Other input and output do not use flow, the flow seems a bit slow.

A better structure on the net (http://www.cnblogs.com/mobileliker/archive/2013/05/26/3099748.html) is to open a sparse array of 10 million to save 0-9999999 of the bar chart count , eliminating the ordering of NLGN. The memory limit is not tight, there are 65536KB, this array shaping words is 40 million bytes, about 35000K bytes, so enough memory.

Another place to learn from this solution is the way he reads it, he reads it in a character, reads it in a single character, looks sloppy, but does save a single read line and iterate the repetition of each character. I/O efficiency is the key

My solution is to learn from the above method, and his solution is also more like, that is true to write is the push this method to a more extreme state.

#include "stdio.h"


int _n, _nodup = 0, _idx[10000000];


int _mapint[25] = {2,2,2,3,3,3,4,4,4,5,5,5,6,6,6,7,0,7,7,8,8,8,9,9,9};


int main () {


Register int _i, _j, _x;


Register int _k;


Register char _c;


scanf ("%d\n", &_n);


for (_i = 0; _i < _n; ++_i) {


_x = 0; _j = 0;


while (_j < 7) {


_c = GetChar ();


if (_c >= ' 0 ' && _c <= ' 9 ') {


_x = _x * + _c-' 0 ';


++_j;


}else if (_c >= ' A ' && _c <= ' Y ' && _c!= ' Q ') {


_x = _x * + _mapint[_c-' A '];


++_j;


            }


        }


_idx[_x]++;


while (GetChar ()!= ' \ n ');


    }





for (_i = 0; _i < 10000000; ++_i) {


_k = _idx[_i];


if (_k > 1) {


printf ("%03d-%04d%d\n", (_i/10000), (_i%10000), _k);


_nodup = 1;


        }


    }





if (_nodup = = 0) printf ("No duplicates.\n");


return 0;


}

Edit 1: Perhaps further acceleration can be performed on the bypass output part of the remainder operator "%".