Baby Ming and phone number
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Time Limit:1500MS
Memory Limit:65536KB 64bit IO Format:%I64d & %I64u
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.
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)
$1 \leq T \leq 30, b < 1000, 0 < a, n \leq 100,000$
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;
}