P7911 [CSP-J 2021] 网络连接 题解

模拟代码

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

int n, p = 1, ans[1003]; // 没事干的ans数组

struct node{
	string op, ad;
}a[1003];

int ws(int n) { // 位数
	int sum = 0;
	if (n == 0) return 1;
	while (n) {
		n /= 10;
		sum++;
	}
	return sum;
}

bool check(string ad) {
	int c1 = 0, c2 = 0;
	for (int i = 0; i < ad.size(); i++) { // 统计 . 和 : 的数量
		if (ad[i] == '.') c1++;
		if (ad[i] == ':') c2++;
	}
	if (c1 != 3 || c2 != 1) return 0;
	c1 = 0, c2 = 0;
	for (int i = 0; i < ad.size(); i++) { // 检查顺序
		if (ad[i] >= '0' && ad[i] <= '9') continue;
		if (ad[i] == '.') c1++;
		if (c1 == 3 && c2 == 0) break; // 当点有 3 个时，冒号肯定还有 1 个，为了避免后面判断错误，所以直接跳出循环
		if ((c1 == 1 || c1 == 2 || c1 == 3) && c2 >= 1) return 0;
	}
	long long num = 0, nump = 1, st = -1, po = 0, cz = 0;
	for (int i = 0; i < ad.size(); i++) {
		if (ad[i] >= '0' && ad[i] <= '9') { // 数字
			num = num * 10 + (ad[i] - '0');
			if (!po) { // 记录起点
				st = i;
				po = 1;	
			}
			cz = 1;
		}	
		else if (nump >= 1 && nump <= 4) {
			if (!cz) return 0;
			if (ws(num) != i - st) return 0; // 位数不符合要求
			if (!(num >= 0 && num <= 255)) return 0;
			num = 0;
			nump++;
			po = 0;
		}
	}
	if (nump != 5) return 0;
	if (ws(num) != ad.size() - st) return 0;
	if (!(num >= 0 && num <= 65535)) return 0; 
	return 1;
}

bool have_server(int n) {
	for (int i = 1; i < n; i++) {
		if (a[i].op == "Server" && a[i].ad == a[n].ad) 
			return 1;
	}
	return 0;
}

int find_server(string ad, int n) {
	for (int i = 1; i < n; i++)
		if (a[i].op == "Server" && a[i].ad == ad) return i;
	return 0;
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		cin >> a[i].op >> a[i].ad;
		if (check(a[i].ad)) {
			if (a[i].op == "Server") {
				if (!have_server(i)) ans[i] = -1;
				else ans[i] = -2;
			}
			if (a[i].op == "Client") {
				int s = find_server(a[i].ad, i);
				if (s) ans[i] = s;
				else ans[i] = -2;
			}
		}
		else ans[i] = -3;
	}
	for (int i = 1; i <= n; i++) {
		if (ans[i] > 0) printf("%d\n", ans[i]);
		else if (ans[i] == -1) printf("OK\n");
		else if (ans[i] == -2) printf("FAIL\n");
		else if (ans[i] == -3) printf("ERR\n");
	}
	return 0;
}