hdoj--5611--Baby Ming and phone number（模拟水题）



Baby Ming and phone number

Crawling in process... Crawling failed Time Limit:1500MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 5611

Description

Baby Ming collected lots of cell phone numbers, and he wants to sell them for money.

He thinks normal number can be sold for $b$ yuan, while number with following features can be sold for $a$ yuan.

1.The last five numbers are the same. (such as 123-4567-7777)

2.The last five numbers are successive increasing or decreasing, and the diffidence between two adjacent digits is $1$. (such as 188-0002-3456)

3.The last eight numbers are a date number, the date of which is between Jar 1st, 1980 and Dec 31th, 2016. (such as 188-1888-0809,means August ninth,1888)

Baby Ming wants to know how much he can earn if he sells all the numbers.

 

Input

In the first line contains a single positive integer $T$, indicating number of test case.

In the second line there is a positive integer $n$, which means how many numbers Baby Ming has.(no two same phone number)

In the third line there are $2$ positive integers $a, b$, which means two kinds of phone number can sell $a$ yuan and $b$ yuan.

In the next $n$ lines there are $n$ cell phone numbers.（|phone number|==11, the first number can’t be 0)

$1 \leq T \leq 30, b < 1000, 0 < a, n \leq 100,000$

 

Output

How much Baby Nero can earn.

 

Sample Input

1
5
100000 1000
12319990212
11111111111
22222223456
10022221111
32165491212 

 

Sample Output

302000 


#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n;
		cin>>n;
		char str[1010];
		__int64 ans=0;
		__int64 a,b;
		scanf("%lld%lld",&a,&b);
		for(int i=0;i<n;i++)
		{
			memset(str,'\0',sizeof(str));
			bool f=false;
			scanf("%s",str);
			for(int i=0;i<11;i++)
			str[i]-='0';
			if(str[10]==str[9]&&str[10]==str[8]&&str[10]==str[7]&&str[10]==str[6])
			f=true;
			else if(str[6]+1==str[7]&&str[7]+1==str[8]&&str[8]+1==str[9]&&str[9]+1==str[10])
			f=true;
			else if(str[6]-1==str[7]&&str[7]-1==str[8]&&str[8]-1==str[9]&&str[9]-1==str[10])
			f=true;
			int year=str[3]*1000+str[4]*100+str[5]*10+str[6];
			int mon=str[7]*10+str[8];
			int day=str[9]*10+str[10];
			if(year>=1980&&year<=2016)
			{
				if(mon==1||mon==3||mon==5||mon==7||mon==8||mon==10||mon==12)
				{
					if(day<=31)
					f=true;
				}
				if(mon==4||mon==6||mon==9||mon==11)
				{
					if(day<=30)
					f=true;
				}
				if(mon==2&&day<=28)
				f=true;
				if((year%4==0&&year%100!=0)||year%400==0)
				if(mon==2&&day==29)
				f=true;
			}
			if(f)
			ans+=a;
			else
			ans+=b;
		}
		printf("%lld\n",ans);
	}
	return 0;
}