Gym - 101190A Abbreviation（模拟）

原题：

ACM ICPC 2016–2017, Northeastern European Regional Contest.
St. Petersburg – Barnaul – Tbilisi – Almaty, December 4, 2016
Problem A. Abbreviation
Input file: abbreviation.in
Output file: abbreviation.out
An abbreviation (from Latin brevis, meaning short) is a shortened form of a word or phrase. In this
problem you must write an automated tool that replaces a sequence of capitalized words with the corresponding abbreviation that consists of the first upper case letters only, followed by a full definition in
parenthesis. See sample input and output.
Let us make some formal definitions. A word in a text is a maximally long sequence of lower and upper
case English letters. A capitalized word is a word that consists of an upper case letter followed by one
or more lower case letters. For example, “Ab”, “Abc”, “Abcd”, and “Abcde“ are all capitalized words,
while “ab”, “A”, “AB“, “ABc“ and “AbC“ are not.
An abbreviatable sequence of words is a sequence of two or more capitalized words that are separated by
exactly one space, no line breaks or punctuation are allowed inside it.
An abbreviation of an abbreviatable sequence of words is a sequence of the first (upper case) letters of
each word, followed by a single space, an opening parenthesis, the original abbreviatable sequence, and
a closing parenthesis.
Input
The input file consists of up to 1 000 lines of text with up to 120 characters on each line. Each line
consists of spaces, upper and lower case letters, commas or dots. There are no leading or trailing spaces
on lines and there are no empty lines. There is at least one line in the input file.
Output
Write to the output file the original text with every abbreviatable sequence of words replaced with the
corresponding abbreviation.
Examples
abbreviation.in
This is ACM North Eastern European Regional Contest,
sponsored by International Business Machines.
The. Best. Contest. Ever.
A Great Opportunity for all contestants.
abbreviation.out
This is ACM NEERC (North Eastern European Regional Contest) ,
sponsored by IBM (International Business Machines).
The. Best. Contest. Ever.
A GO (Great Opportunity) for all contestants.
abbreviation.in
ab Ab A Abc AB Abcd ABc Abcde AbC
abbreviation.out
ab Ab A Abc AB Abcd ABc Abcde AbC
abbreviation.in
Oh No Extra Spaces.And,Punctuation Ruin Everything
abbreviation.out
Oh No ES (Extra Spaces).And,PRE (Punctuation Ruin Everything)



题意：

       给出一串字符，按要求将一部分改为缩写

按题意模拟就好

#include <iostream>
#include <iomanip>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <deque>
#include <string>
#include <cmath>
#include <vector>
#include <utility>
#include <set>
#include <map>
#include <sstream>
#include <climits>
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#define pi acos(-1.0)
#define INF 2147483647
using namespace std;
typedef long long ll;
typedef pair <int,int > P;
char s[130];
int main ()
{
    freopen("abbreviation.in", "r", stdin);
    freopen("abbreviation.out", "w", stdout);
    while(gets(s))
    {
        int len=strlen(s);
        string suo,yuan,ss;
        for(int i=0; i<len; )
        {
            string ab,qq;
            int big=0,little=0;
            bool ok=false;
            while((s[i]>='a'&&s[i]<='z')||(s[i]>='A'&&s[i]<='Z')&&i<len)
            {
                qq+=s[i];
                ok=true;
                ss+=s[i];
                if(s[i]>='A'&&s[i]<='Z')
                    big++;
                else
                    little++;
                ab+=s[i];
                i++;
            }
            int num1=0,num2=0;
            bool ins=false;
            while(!(s[i]>='a'&&s[i]<='z')&&!(s[i]>='A'&&s[i]<='Z')&&i<len)
            {
                qq+=s[i];
                if(s[i]!=' ')
                {
                    num2++;
                    ins=true;
                }
                ok=true;
                ss+=s[i];
                if(s[i]==' ')
                    num1++;
                i++;
            }
            if((ab[0]>='A'&&ab[0]<='Z')&&big==1&&little>=1)
            {
                suo+=ab[0];
                yuan+=ab;
            }
            else
            {
                if(suo.size()>1)
                {
                    int len1=yuan.size();
                    cout<<suo<<" (";
                    for(int j=0; j<len1-1; j++)
                    {
                        cout<<yuan[j];
                    }
                    cout<<")";
                    cout<<" "<<qq;
                }
                else
                    cout<<ss;
                ss.clear();
                suo.clear();
                yuan.clear();
            }
            if(num1>1||ins||i==len)
            {
                if(suo.size()>1)
                {
                    cout<<suo<<" (";
                    for(int j=0; j<yuan.size(); j++)
                    {
                        cout<<yuan[j];
                    }
                    cout<<")";
                    for(int k=num1+num2; k>0; k--)
                        cout<<s[i-k];
                }
                else
                    cout<<ss;
                ss.clear();
                suo.clear();
                yuan.clear();
            }
            else if(num1==1)
                yuan+=s[i-1];
            if(!suo.size())
                yuan.clear();
        }
        cout<<endl;
    }
    return 0;
}