题解

司马Zzz超我题解
CSPJ都这么难的吗
超绝大模拟
有几个坑点注意一下：
1.不允许前导0不代表不允许地址的某一个地方为0，比如说 192.168.0.255:80 中第三个区域一开始是0，但是它是合法的
2.要判断一开始是不是数字，比如 .123.213.231:14514 不是合法的，很难找到这个问题
来给你们造个最全的数据

8
Client 0.0.0.0:0
Client 255.255.255.255:65535
Client .0.0.0:0
Client 0.0.0.0:
Client 0.0.0.0:0:
Client 2147483649.2147483649.2147483649.2147483649:2147483649
Client 0.0.00.0:0
Client 0.0..0:0

FAIL
FAIL
ERR
ERR
ERR
ERR
ERR
ERR

只要判断好错误，剩下的就很简单了
用一个vector储存记录，循环判断重复

#include <iostream>
#include <memory.h> 
#include <cstdio>
#include <vector>
#include <set>
using namespace std;
struct node{
	int a[6] = {0};
};
vector <node> v;
int a[100005];
char b[100005];
int main(){
	int t, __ct = 0;
	cin >> t;
	if(0){error: cout << "ERR\n";}
	if(0){fail: cout << "FAIL\n";}
	while(t--){
        __ct++;
        
		memset(a, 0, 1000);
		memset(b, 0, 1000);
		string tmp, str;
		cin >> tmp >> str;
		int ct = 1, i;
		if(!isdigit(str[0])) goto error;
		for(i = 0; str[i] != '\0'; i++){
			if(!isdigit(str[i])){
				if(ct <= 4 && a[ct] > 255 || a[ct] > 65535 || i && !isdigit(str[i - 1])) goto error;
				b[ct++] = str[i];
			}
			else{
				if(a[ct] == 0 && str[i] == '0' && isdigit(str[i + 1])) goto error;
				a[ct] = a[ct] * 10 + str[i] - '0';
                if(ct <= 4 && a[ct] > 255 || a[ct] > 65535) goto error;
			}
		}
		if(ct != 5 || a[5] > 65535 || !isdigit(str[i - 1])) goto error;
		for(int i = 1; i <= 3; i++){
			if(b[i] != '.') goto error;
		}
		if(b[4] != ':') goto error;
		
		if(tmp == "Server"){
			for(auto it:v){
				bool flag = 1;
				for(int i = 1; i <= 5; i++) if(it.a[i] != a[i]) flag = 0;
				if(flag) goto fail;
			}
			cout << "OK\n";
			node tmp;
			for(int i = 1; i <= 5; i++) tmp.a[i] = a[i];
            tmp.a[0] = __ct;
			v.push_back(tmp);
		}else{
			int idx = -1;
			for(int i = 0; i < v.size(); i++){
				bool flag = 1;
				for(int j = 1; j <= 5; j++){
					if(v[i].a[j] != a[j]) flag = 0;
				}
				if(flag){
					idx = v[i].a[0];
					break;
				}
			}
			if(idx == -1) goto fail;
			cout << idx << endl;
		}
	}

	return 0;
}