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



Baby Ming and phone number

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Time Limit:1500MS    
Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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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;
}