BestCoder Round #69 (div.2) B HDOJ5611 Baby Ming and phone number(stl)

Baby Ming and phone number

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 373    Accepted Submission(s): 108

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
  
  

 

题目链接:点击打开链接

n个电话号码, 满足条件的可以卖a元, 其他号码卖b元, 计算可以卖的钱数. 条件: 1.后五位相等 2.后五位递增或递减 3.后8位是正确的日

期(1980.1.1 - 2016.12.31).

1 2条件容易判断, 第3个条件利用stl string的assign()处理, 直接比较即可判断.

AC代码:

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "queue"
#include "stack"
#include "cmath"
#include "utility"
#include "map"
#include "set"
#include "vector"
#include "list"
#include "string"
#include "cstdlib"
using namespace std;
typedef long long ll;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e5 + 5;
int t, n, a, b;
string s;
bool Judge(string x)
{
	int num = 0, cnt = 1;
	for(int i = 3; i >= 0; --i) {
		num += cnt * x[i];
		cnt *= 10;
	}
	if(num % 400 == 0 || (num % 4 == 0 && num % 100 != 0)) return true;
	return false;
}
int main(int argc, char const *argv[])
{
	scanf("%d", &t);
	while(t--) {
		int num1 = 0, num2 = 0;
		scanf("%d", &n);
		scanf("%d%d", &a, &b);
		for(int i = 0; i < n; ++i) {
			cin >> s;
			string y, m, d;
			y.assign(s, 3, 4);
			m.assign(s, 7, 2);
			d.assign(s, 9, 2);
			if(s[6] == s[7] && s[7] == s[8] && s[8] == s[9] && s[9] == s[10]) num1++;
			else if(s[6] + 1 == s[7] && s[7] + 1 == s[8] && s[8] + 1 == s[9] && s[9] + 1 == s[10]) num1++;
			else if(s[6] - 1 == s[7] && s[7] - 1 == s[8] && s[8] - 1 == s[9] && s[9] - 1 == s[10]) num1++;
			else if(y >= "1980" && y <= "2016") {
				if(m == "01" && d <= "31" && d >= "01") num1++;
				else if(m == "03" && d <= "31" && d >= "01") num1++;
				else if(m == "05" && d <= "31" && d >= "01") num1++;
				else if(m == "07" && d <= "31" && d >= "01") num1++;
				else if(m == "08" && d <= "31" && d >= "01") num1++;
				else if(m == "10" && d <= "31" && d >= "01") num1++;
				else if(m == "12" && d <= "31" && d >= "01") num1++;
				else if(m == "04" && d <= "30" && d >= "01") num1++;
				else if(m == "06" && d <= "30" && d >= "01") num1++;
				else if(m == "09" && d <= "30" && d >= "01") num1++;
				else if(m == "11" && d <= "30" && d >= "01") num1++;
				else if(m == "02" && ((Judge(y) && d <= "29" && d >= "01") || (!Judge(y) && d <= "28" && d >= "01"))) num1++;
				else num2++;
			}
			else num2++;
		}
		printf("%lld\n", 1ll * num1 * a + 1ll * num2 * b);
	}
	return 0;
}