HDU-5611 Baby Ming and phone number（模拟）

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<a,n≤100,000

 

Output

How much Baby Nero can earn.

 

Sample Input

15100000 10001231999021211111111111222222234561002222111132165491212

 

Sample Output

302000

水题，按照题意判断即可

忘了0开头的数字表示8进制，01234导致数字错误，浪费好长时间

#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\n",ans);    }    return 0;}