codeforce727b---Bill total Value (string-handling regular expression)

Let's start with the regular expression.

%*[a-z] means ignoring the preceding lowercase characters,%[0-9] means to deposit the contiguous number of non-characters into the T string;

Get "123" from "abc123de4f";

#include <stdio.h>int main () {    char s[50] = "abc123de4f", t[50] = {0};    SSCANF (S, "%*[a-z]%[0-9]", t);    printf ("%s\n", t);//-Result: 123    return 0;}

Get "12.3" from "abc12.3de4f"

#include <stdio.h>int main () {    char s[50] = "abc12.3de4f", t[50] = {0};    SSCANF (S, "%*[^0-9]%[0-9.]", t);    printf ("%s\n", t);//-Result: 12.3    return 0;}

　　

Get "@qq. com" from "[email protected]"

#include <stdio.h>int main () {    char s[50] = "[email protected]", t[50] = {0};    SSCANF (S, "%*[^@]%s", t);    printf ("%s\n", t); results: @qq. com    return 0;}

　　Get "qq.com" from "[email protected]"

#include <stdio.h>int main () {    char s[50] = "[email protected]", t[50] = {0};    SSCANF (S, "%*[^@]@%s", t);    printf ("%s\n", t);//-Result: 12.3    return 0;}

There are a lot of things you can try slowly.

Title Link: Http://codeforces.com/contest/727/problem/B

Test instructions: There is a purchase order in the form of a string, in order to include the "Item name + Price", there is no "+" number, the item name is composed of lowercase English characters, the price has two forms, with cents and no cent, of which with cent is a decimal point plus two digits of the structure

, the dollar part is divided by the decimal point every 3 bits, for example, 123.45 for $123.45, 12.345 for $12345, and 1.234.45 for $1234.45, which now requires the sum of the prices of all items

The output format is processed according to the input format;

There are two types of output formats:

1. All integers, no. 00 this form

2. There are fractional parts, the fractional part is not 0 reserved two decimal places;

#include <stdio.h>#include<string.h>#defineN 5200intMain () {Chars[n]={0}, next[n]={0}, p[n]={0}, T[n] = {0};  while(SCANF ("%s", s+1) !=EOF) {s[0] ='a';///The rules for the following regular expressions are the beginning of a character, ending with a number;        Doublesum =0; intf =0; ///The regular expression asks the number part in the s string to be stored in the form of a string in P, and the remainder is deposited in next; ///         while(SSCANF (s),"%*[^0-9]%[0-9.] %[^\0]", p, next) >0)        {            DoubleNUM1 =0, num2 =0;///NUM1 is the value of fractional parts, and num2 is the value of the integer part;            intlen = strlen (p), flag =0; if(len>3&& p[len-3] =='.') {sscanf (P+len-3,"%LF", &AMP;NUM1);///converts a decimal part into a double;Flag = f =1;///description is 2 decimal places;            }            if(Flag = =1) len = len-3;  for(intI=0; i<len; i++)///processing integer parts;            {                if(P[i] = ='.')Continue; Num2= num2*Ten+ (p[i]-'0'); } Sum+ = Num1 +num2; strcpy (S, next);///cyclic processing of the remaining parts;Memset (Next,0,sizeof(next));///to be emptied;} sprintf (T,"%.2f", sum);///And then save the result as a string to facilitate the output;        intLen = strlen (t), k =0; CharAns[n];///Save the results;        if(f = =0) Len-=3;///having no part of a small number;        if(f = =1)///2 decimal places;{ans[k+ +] = t[--Len]; Ans[k+ +] = t[--Len]; Ans[k+ +] = t[--Len]; if(ans[0] =='0'&& ans[1] =='0') K =0;///the decimal part is divided into 0, not output;        }         for(inti=len-1, counts=1; i>=0; I--, counts++)        {            if(counts%3==1&& counts!=1) Ans[k++] ='.'; Ans[k++] =T[i]; }         for(inti=k-1; i>=0; i--) printf ("%c", Ans[i]); printf ("\ n"); }    return 0;}

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codeforce727b---Bill total Value (string-handling regular expression)