HDOJ 5611 Baby Ming and phone number(シミュレーション)
4849 ワード
Baby Ming and phone number
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 222 Accepted Submission(s): 68
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
302000
タイトル:
考え方:直接シミュレーションすればいい、月と日付の判断に注意するだけで、結果が爆発する.
acコード:
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 222 Accepted Submission(s): 68
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
タイトル:
， ， 。
a ， ， b 。
1. 5 （ 123-4567-7777）
2. 5 ， 1 （ 188-0002-3456）
3. 8 ， 1980 1 1 2016 12 31 （ 188-1888-0809 1888 8 9 ）
。考え方:直接シミュレーションすればいい、月と日付の判断に注意するだけで、結果が爆発する.
acコード:
#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1010000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
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';
int bz=0;
if(s[10]==s[9]&&s[9]==s[8]&&s[8]==s[7]&&s[7]==s[6])
ans+=a,bz=1;
//printf("bz=%d
",bz);
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]))&&bz==0)
{
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,bz=1;
}
//printf("bz=%d
",bz);
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];
//printf("y=%d m=%d d=%d
",year,month,day);
if(check(year,month,day)&&bz==0)
ans+=a,bz=1;
if(bz==0)
ans+=b;
}
printf("%I64d
",ans);
}
return 0;
}