hdoj5611Baby Ming and phone number

Baby Ming and phone number

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 685    Accepted Submission(s): 202

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

 

Source

BestCoder Round #69 (div.2)

 

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#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<map>#include<queue>using namespace std;const int maxn=10010;char str[15];int mov[15]={0,31,28,31,30,31,30,31,31,30,31,30,31};int main(){    int t,i,j,k,n;    long long a,b,c,ans;    scanf("%d",&t);    while(t--){        scanf("%d%lld%lld",&n,&a,&b);ans=0;        for(j=0;j<n;++j){            scanf("%s",str+1);            bool flag1=true,flag2=true,flag3=true;            for(i=7;i<=11;++i){                if(str[i]!=str[7])flag1=false;                if((str[i]-str[7])!=(i-7))flag2=false;                if((str[i]-str[7])!=(7-i))flag3=false;            }            int y=(str[4]-'0')*1000+(str[5]-'0')*100+(str[6]-'0')*10+(str[7]-'0');            int m=(str[8]-'0')*10+str[9]-'0';            int d=(str[10]-'0')*10+str[11]-'0';            bool flag4=false;            if(y>=1980&&y<=2016){                if(m>=1&&m<=12){                    if((y%4==0&&y%100!=0)||y%400==0){                        mov[2]++;                        if(d>=1&&d<=mov[m])flag4=true;                        mov[2]--;                    }                    else if(d>=1&&d<=mov[m])flag4=true;                }            }                        if(flag1||flag2||flag3||flag4)ans+=a;            else ans+=b;        }        printf("%lld\n",ans);    }    return 0;}