bzoj4839 [Neerc2016]Abbreviation

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=4839

【题解】

pkusc怎么出bzoj原题啊

字符串随便处理一下就行了，大模拟

pkusc：2A（freopen调试没删）

线下：2A（数组不够大）

# include <stdio.h>
# include <string.h>
# include <iostream>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 5e5 + ;
const int mod = 1e9+;

# define RG register
# define ST static

char str[]; 

char s[][];
int sn = ;
char t[];
char ans[];
int tn = , n = ; 

inline bool isok(int i) {
    if(!isupper(s[i][])) return ;
    int len = strlen(s[i]);
    if(len == ) return ;
    for (int j=; j<len; ++j)
        if(!islower(s[i][j])) return ;
    return ;
}

inline bool isBlank(int i) {
    int len = strlen(s[i]);
    if(len != ) return ;
    return s[i][] == ' ';
}

int main() {
//    freopen("bzoj4839.in", "r", stdin);
    while(cin.getline(str, )) {
        int len = strlen(str);
        sn = ;
        for (int i=; i<len; ++i) {
            if(isupper(str[i]) || islower(str[i])) {
                tn = ;
                int j = i;
                while(isupper(str[j]) || islower(str[j])) {
                    t[tn++] = str[j];
                    ++j;
                }
                ++sn;
                for (int j=; j<tn; ++j)
                    s[sn][j] = t[j];
                s[sn][tn] = ;
                i = j-;
            } else {
                ++sn; s[sn][] = str[i];
                s[sn][] = ;
            }
        }
//        for (int i=1; i<=sn; ++i) printf("%s==\n", s[i]);
        n = ;
        for (int i=; i<=sn; ++i) {
            if(!isok(i)) {
                for (int j=; s[i][j]; ++j)
                    ans[n++] = s[i][j];
            } else {
                int j = i;
                while(j+ <= sn && isBlank(j+) && isok(j+)) j += ;
                if(j == i) {
                    for (int k=; s[i][k]; ++k)
                        ans[n++] = s[i][k];
                } else {
                     for (int k=i; k<=j; k+=)
                         ans[n++] = s[k][];
                    ans[n++] = ' ';
                    ans[n++] = '(';
                    for (int k=i; k<=j; k++)
                        for (int l=; s[k][l]; ++l)
                            ans[n++] = s[k][l];
                    ans[n++] = ')';
                    i = j;
                }
            }
        }
        ans[n] = ;
        printf("%s\n", ans);
    }
    return ;
}