从后台研发到跑路

// Problem: 从后台研发到跑路
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/20960/1006
// Memory Limit: 262144 MB
// Time Limit: 2000 ms
// 2022-03-24 01:38:30
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;   
typedef vector<ll> Vll;               
typedef vector<pair<int, int> > vpii;
typedef vector<pair<ll, ll> > vpll;                        

const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const double pi  = acos(-1.0);

inline ll qmi (ll a, ll b) {
	ll ans = 1;
	while (b) {
		if (b & 1) ans = ans * a%mod;
		a = a * a %mod;
		b >>= 1;
	}
	return ans;
}
inline int read () {
	int x = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
	while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
	return f?-x:x;
}
template<typename T> void print(T x) {
	if (x < 0) putchar('-'), x = -x;
	if (x >= 10) print(x/10);
	putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
	return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
	return (a + b) %mod;
}
inline ll inv (ll a) {
	return qmi(a, mod - 2);
}

char s[100000];
void solve() {
	char x;
	int n = 0;
	while((x = getchar()) != EOF) {
		s[n++] = x;
	}
	for (int i = 0; i < n; i ++) {
		if(s[i] == '\"' && s[i  -1] != '\\') {
			cout << s[i++];
			while (1) {
				cout << s[i];
				if (s[i] == '"' && s[i - 1] != '\\')
					break;
				i ++;
			}
		}
		else if (s[i] == '/' && s[i + 1] == '*') {
			int j = i;
			i += 3;
			while (1) {
				if (i >= n || s[i - 1] == '*' && s[i] == '/')
					 break;
				i ++;
			}
			if (i >= n)
				for (;j < n;j ++)
					cout << s[j];
		}
		else cout << s[i];
	}
	
}
int main () {
	// ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);
    int t;
    t =1;
    //cin >> t;
    while (t --) solve();
    return 0;
}

我们从终止条件的角度处理