Hdu 5611 Baby Ming and phone number【日期计算等等】

Baby Ming and phone number

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 1656 Accepted Submission(s): 434

[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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

[align=left]Output[/align]
How much Baby Nero can earn.

[align=left]Sample Input[/align]

1
5
100000 1000
12319990212
11111111111
22222223456
10022221111
32165491212

[align=left]Sample Output[/align]

302000

题意：

给出一定的条件，满足条件的可以卖a元，否则b元，问这个人能赚多少钱。

题解：

需要完全模拟题目的条件，另外日期计算的时候注意细节的考虑，比如闰年等

#include<stdio.h>
typedef long long ll;
int x[]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int abs(int x)
{
return x<0?-x:x;
}
int fun(int x)
{
return x%4==0&&x%100!=0||x%400==0;
}
bool one(char s[])
{
for(int i=7;i<11;++i)
{
if(s[i]!=s[i-1])
{
return 0;
}
}
return 1;
}
bool two(char s[])
{
for(int i=7;i<10;++i)
{
if(s[i]-s[i-1]!=s[i+1]-s[i]||abs(s[i]-s[i-1])!=1)
{
return 0;
}
}
return 1;
}
bool three(char s[])
{
int year=0,month=0,day=0;
for(int i=3;i<11;++i)
{
if(i<7)
{
year=year*10+(s[i]-'0');
}
else if(i<9)
{
month=month*10+(s[i]-'0');
}
else
{
day=day*10+(s[i]-'0');
}
}
if(year<1980||year>2016||month==0||month>12||day==0||day>31)
//不符合日期的规则
{
return 0;
}
if(month==2)//讨论二月
{
int tp=fun(year);//判断闰年
return day<=x[month]+tp;
}
else//其他月份
{
return day<=x[month];
}
}
bool judge(char s[])
{
return one(s)||two(s)||three(s);
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
ll n,a,b;
scanf("%I64d%I64d%I64d",&n,&a,&b);
ll ans=0;char s[15];
while(n--)
{
scanf("%s",s);
ans+=judge(s)?a:b;
}
printf("%I64d\n",ans);
}
return 0;
}