BestCoder Round #69 (div.2)（hdu5611）

Posted 2020-06-17

Baby Ming and phone number

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1501    Accepted Submission(s): 399

Problem 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≤T≤30,b<1000,0<a,n≤100,000

 

 

Output

How much Baby Nero can earn.

 

 

Sample Input

1

5

100000 1000

12319990212

11111111111

22222223456

10022221111

32165491212

 

 

Sample Output

302000

 

分步判断是否符合题意即可   休要细心，代码未ac  还有细节未处理好  先记录下来

#include<stdio.h>
#include<string.h>
#include<string>
#include<math.h>
#include<algorithm>
#define LL long long
#define PI atan(1.0)*4
#define DD doublea
#define MAX 100100
#define mod 10007
using namespace std;
int s[MAX][15];
char s1[MAX][15];
int op1[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int op2[13]={0,31,29,31,30,31,30,31,31,30,31,30,31};
int five(int x)//判断后五位 
{
	int i,j;
	char y=s[x][6];
	int flag=1;
	for(i=7;i<11;i++)
	{
		if(s[x][i]!=y)
		{
			flag=0;
			break;
		}
	}
	if(flag==1) return 1;
	flag=1;
	for(i=7;i<11;i++)
	{
		if(s[x][i]!=s[x][i-1]+1)
		{
			flag=0;
			break;
		}
	}
	if(flag==1) return 1;
	else return 0;
}
int judge(int x)//判断闰年 
{
	if((x%4==0&&x%100!=0)||x%400==0)
	    return 1;
	else return 0;
}
int eight(int x)  //判断后八位 
{
	int i,j;
	int flag=1;
	int sum=s[x][3];
	int mouth=s[x][7]*10+s[x][8];
	int day=s[x][9]*10+s[x][10];
	for(i=4;i<=6;i++)
 	   sum=sum*10+s[x][i];
 	if(sum<1980||sum>2016||mouth<1||mouth>12)
 	    return 0;
 	if(sum==1980)
 	{
 		if(mouth<7||mouth>12)
 		    return 0;
 	    if(day==0||day>op1[mouth])
 	        return 0;
 	}
 	if(judge(sum))
 	{
 	    if(day==0||day>op2[mouth])
 	        return 0;	
 	}
 	if(!judge(sum))
 	{
 		if(day==0||day>op1[mouth])
 	        return 0;
 	}
 	return 1;
}
int main()
{
    int t,i,j,n,m;
    LL ans,a,b;
	scanf("%d",&t);	
	while(t--)
	{
		ans=0;
		scanf("%d",&n);
		scanf("%lld%lld",&a,&b);
		for(i=1;i<=n;i++)
		{
			scanf("%s",s1[i]);
			for(j=0;j<11;j++)
			{
				s[i][j]=s1[i][j]-‘0‘;
			}
		}
		for(i=1;i<=n;i++)
		{
			if(five(i))
				ans+=a;
			else if(eight(i))
				ans+=a;
			else ans+=b;
		}
		printf("%lld\n",ans);
	}
	return 0;
}