Baby Ming and phone number
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1501 Accepted Submission(s): 399
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.
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
#include<stdio.h>
#include<string.h>
#include<string>
#include<math.h>
#include<algorithm>
#define LL long long
#define PI atan(1.0)*4
#define DD doublea
#define MAX 100100
#define mod 10007
using namespace std;
int s[MAX][15];
char s1[MAX][15];
int op1[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int op2[13]={0,31,29,31,30,31,30,31,31,30,31,30,31};
int five(int x)//判断后五位
{
int i,j;
char y=s[x][6];
int flag=1;
for(i=7;i<11;i++)
{
if(s[x][i]!=y)
{
flag=0;
break;
}
}
if(flag==1) return 1;
flag=1;
for(i=7;i<11;i++)
{
if(s[x][i]!=s[x][i-1]+1)
{
flag=0;
break;
}
}
if(flag==1) return 1;
else return 0;
}
int judge(int x)//判断闰年
{
if((x%4==0&&x%100!=0)||x%400==0)
return 1;
else return 0;
}
int eight(int x) //判断后八位
{
int i,j;
int flag=1;
int sum=s[x][3];
int mouth=s[x][7]*10+s[x][8];
int day=s[x][9]*10+s[x][10];
for(i=4;i<=6;i++)
sum=sum*10+s[x][i];
if(sum<1980||sum>2016||mouth<1||mouth>12)
return 0;
if(sum==1980)
{
if(mouth<7||mouth>12)
return 0;
if(day==0||day>op1[mouth])
return 0;
}
if(judge(sum))
{
if(day==0||day>op2[mouth])
return 0;
}
if(!judge(sum))
{
if(day==0||day>op1[mouth])
return 0;
}
return 1;
}
int main()
{
int t,i,j,n,m;
LL ans,a,b;
scanf("%d",&t);
while(t--)
{
ans=0;
scanf("%d",&n);
scanf("%lld%lld",&a,&b);
for(i=1;i<=n;i++)
{
scanf("%s",s1[i]);
for(j=0;j<11;j++)
{
s[i][j]=s1[i][j]-'0';
}
}
for(i=1;i<=n;i++)
{
if(five(i))
ans+=a;
else if(eight(i))
ans+=a;
else ans+=b;
}
printf("%lld\n",ans);
}
return 0;
}
今天又仔细看了这道题，发现昨天少考虑一种情况，题目有三个要求1、后五位数字一样 2、后五位数字连续（可以递增也可以递减 忘记考虑递减的情况了）3、后八位可以组成一个日期，日期范围是1980年01月01号到2016年12月31号
接下来就主要处理每个月份对应时间以及闰年即可
#include<stdio.h>
#include<string.h>
#include<string>
#include<math.h>
#include<algorithm>
#define LL long long
#define PI atan(1.0)*4
#define DD doublea
#define MAX 100100
#define mod 10007
using namespace std;
int s[MAX][15];
char s1[MAX][15];
int op[13]={0,31,0,31,30,31,30,31,31,30,31,30,31};
//int op2[13]={0,31,29,31,30,31,30,31,31,30,31,30,31};
int five(int x)//判断后五位
{
int i,j;
int y=s[x][6];
int flag=1;
for(i=7;i<11;i++)
{
if(s[x][i]!=y)
{
flag=0;
break;
}
}
if(flag==1) return 1;
flag=1;
for(i=7;i<11;i++)
{
if(s[x][i]!=s[x][i-1]+1)
{
flag=0;
break;
}
}
if(flag==1) return 1;
flag=1;
for(i=7;i<11;i++)
{
if(s[x][i]!=s[x][i-1]-1)
{
flag=0;
break;
}
}
if(flag==1) return 1;
else return 0;
}
int judge(int x)//判断闰年
{
if((x%4==0&&x%100!=0)||x%400==0)
return 1;
else return 0;
}
int eight(int x) //判断后八位
{
int i,j;
int flag=1;
int sum=s[x][3];
int mouth=s[x][7]*10+s[x][8];
int day=s[x][9]*10+s[x][10]; for(i=4;i<=6;i++)
sum=sum*10+s[x][i];
// printf("%d*\n",sum);
if(mouth==2)
{
if(judge(sum))
op[2]=29;
else
op[2]=28;
}
if(sum>=1980&&sum<=2016&&mouth>=1&&mouth<=12&&day>=1&&day<=op[mouth])
return 1;
else return 0;
}
int main()
{
int t,i,j,n,m;
__int64 ans,a,b;
scanf("%d",&t);
while(t--)
{
ans=0;
scanf("%d",&n);
scanf("%I64d%I64d",&a,&b);
for(i=1;i<=n;i++)
{
scanf("%s",s1[i]);
for(j=0;j<11;j++)
s[i][j]=s1[i][j]-'0';
}
for(i=1;i<=n;i++)
{
if(five(i))
ans+=a;
else if(eight(i))
ans+=a;
else ans+=b;
}
printf("%I64d\n",ans);
}
return 0;
}