HDOJ 5611-Baby Ming and phone number【模拟】

超、凢脫俗 2022-08-20 04:10 23阅读 0赞

# Baby Ming and phone number #

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

Total Submission(s): 958 Accepted Submission(s): 282

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![Image 1][] yuan, while number with following features can be sold for  a![Image 1][] 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![Image 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![Image 1][], indicating number of test case.  
  
In the second line there is a positive integer  n![Image 1][], which means how many numbers Baby Ming has.(no two same phone number)  
  
In the third line there are  2![Image 1][] positive integers  a,b![Image 1][], which means two kinds of phone number can sell  a![Image 1][] yuan and  b![Image 1][] yuan.  
  
In the next  n![Image 1][] lines there are  n![Image 1][] cell phone numbers.（|phone number|==11, the first number can’t be 0)  
  
 1≤T≤30,b<1000,0<a,n≤100,000![Image 1][]

Output

How much Baby Nero can earn.

Sample Input

1
    5
    100000 1000
    12319990212
    11111111111
    22222223456
    10022221111
    32165491212

Sample Output

302000

Source

[BestCoder Round \#69 (div.2) ][BestCoder Round _69 _div.2_]

解题思路：

题目大意就是：

铭宝宝收集了很多手机号码，没错，他想卖手机号码赚钱。
    他觉得有如下性质的手机号码可以卖aaa元钱，其他的号码，只能卖bbb元钱。
    1.末5位数字相同（比如123-4567-7777）
    2.末5位是连续递增或者连续递减的，且相邻数位相差1的数（比如188-0002-3456）
    3.末8位是一个表示日期的数字，并且表示的日期在1980年1月1日至2016年12月31日内（比如188-1888-0809表示1888年8月9日）
    铭宝宝想知道卖掉所有的手机号码能赚多少钱。

输入T(T≤30)T(T \leq 30)T(T≤30)表示TTT组测试数据
    输入n(n≤100,000)n(n \leq 100,000)n(n≤100,000)表示铭宝宝有nnn张手机号码（没有相同的手机号码）
    输入222个正整数a,ba, ba,b, 表示两种类型的手机号码分别能卖aaa元和bbb元(b≤1000,a≤100,000)(b \leq 1000, a \leq 100,000)(b≤1000,a≤100,000)
    接下去nnn行，每行输入111个手机号码（|phonenumber|==11，首位非0）

所以简单模拟就行了，注意最后的结果会超int.

#include<stdio.h>
    #include<string.h>
    #include<algorithm>
    #define ll long long 
    using namespace std;  
    char str[15];  
    int s[15];  
    int m[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};  
    int check(int a,int b,int c)  
    {  
        if(a<1980||a>2016)  
        return 0;  
        if(b<1||b>12)  
        return 0;  
        if(c<1||c>31)  
        return 0;  
        if(a%400==0||(a%4==0&&a%100!=0))  
        {  
            if(b==2)  
            {  
                if(c<=29)  
                return 1;  
                else  
                return 0;  
            }  
            else  
            {  
                if(c>m[b])  
                return 0;  
                else  
                return 1;  
            }  
        }  
        else  
        {  
            if(b==2)  
            {  
                if(c<=28)  
                return 1;  
                else  
                return 0;  
            }  
            else  
            {  
                if(c>m[b])  
                return 0;  
                else  
                return 1;  
            }  
        }  
    }  
    int main()  
    {  
        int t,a,b,c;  
        int i,j,n;  
        scanf("%d",&t);  
        while(t--)  
        {  
            scanf("%d",&n);  
            scanf("%d%d",&a,&b);  
            ll ans=0;  
            while(n--)  
            {  
                scanf("%s",str);  
                int len=strlen(str);  
                for(i=0;i<len;i++)  
                s[i]=str[i]-'0';   
                if(s[10]==s[9]&&s[9]==s[8]&&s[8]==s[7]&&s[7]==s[6])
    			{
    				ans+=a; 
    				continue;
    			}  
                if((s[10]>s[9]&&s[9]>s[8]&&s[8]>s[7]&&s[7]>s[6])||(s[10]<s[9]&&s[9]<s[8]&&s[8]<s[7]&&s[7]<s[6])) 
                {  
                    if(abs(s[10]-s[9])==1&&abs(s[9]-s[8])==1&&abs(s[8]-s[7])==1&&abs(s[7]-s[6])==1)  
                    {
                    	ans+=a;
                    	continue;
                    }
                }  
                int year=s[3]*1000+s[4]*100+s[5]*10+s[6];  
                int month=s[7]*10+s[8];  
                int day=s[9]*10+s[10];
                if(check(year,month,day))
                {
                	ans+=a;
                	continue;
                }  
                ans+=b;  
            }  
            printf("%I64d\n",ans);  
        }  
        return 0;  
    }

[Image 1]: 
[BestCoder Round _69 _div.2_]: http://acm.hdu.edu.cn/search.php?field=problem&key=BestCoder+Round+%2369+%28div.2%29&source=1&searchmode=source