HDU-5611 Baby Ming and phone number(アナログ)
Baby Ming and phone number
http://acm.hdu.edu.cn/showproblem.php?pid=5611 Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
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
Output
How much Baby Nero can earn.
Sample Input
Sample Output
水の問題は、問題の意味によって判断すればいい.
0の先頭の数字を忘れて8進数を表して、01234数字の誤りを招いて、とても長い時間を浪費します
http://acm.hdu.edu.cn/showproblem.php?pid=5611 Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
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
Output
How much Baby Nero can earn.
Sample Input
1
5
100000 1000
12319990212
11111111111
22222223456
10022221111
32165491212
Sample Output
302000
水の問題は、問題の意味によって判断すればいい.
0の先頭の数字を忘れて8進数を表して、01234数字の誤りを招いて、とても長い時間を浪費します
#include <cstdio>
using namespace std;
const int days[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
const int fea[22]={ 0,11111,22222,33333,44444,55555,66666,77777,88888,99999,
1234,12345,23456,34567,45678,56789,
98765,87654,76543,65432,54321,43210
};
int T,n;
long long number,ans,a,b;
inline bool isSpic(int num) {
for(int i=0;i<22;++i)
if(num==fea[i])
return true;
return false;
}
inline int isleap(int year) {
if((year%4==0&&year%100!=0)||year%400==0)
return 1;
return 0;
}
bool isDate(int year,int month,int day) {
if(year<1980||year>2016||month==0||month>12||day==0)
return false;
if(month==2) {
if(day>28+isleap(year))
return false;
return true;
}
if(day>days[month])
return false;
return true;
}
int main() {
scanf("%d",&T);
while(T--) {
ans=0;
scanf("%d%I64d%I64d",&n,&a,&b);
while(n--) {
scanf("%I64d",&number);
ans+=(isSpic(number%100000)||isDate((number/10000)%10000,(number/100)%100,number%100))?a:b;
}
printf("%I64d
",ans);
}
return 0;
}