upc组队赛3 Congestion Charging Zon【模拟】

Congestion Charging Zon

题目描述

Tehran municipality has set up a new charging method for the Congestion Charging Zone (CCZ) which controls the passage of vehicles in Tehran’s high-congestion areas in the congestion period (CP) from 6:30 to 19:00. There are plate detection cameras inside or at the entrances of the CCZ recording vehicles seen at the CCZ. The table below summarizes the new charging method.

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Note that the first time and the last time that a vehicle is seen in the CP may be the same. Write a program to compute the amount of charge of a given vehicle in a specific day.

输入

The first line of the input contains a positive integer n (1 ⩽ n ⩽ 100) where n is the number of records for a vehicle. Each of the next n lines contains a time at which the vehicle is seen. Each time is of form :, where is an integer number between 0 and 23 (inclusive) and is formatted as an exactly two-digit number between 00 and 59 (inclusive).

输出

Print the charge to be paid by the owner of the vehicle in the output.

样例输入

4
7:30
2:20
7:30
17:30

样例输出

36000

题解

大模拟

代码

#include<bits/stdc++.h>
using namespace std;
#define rep(i,a,n) for(int i=a;i<n;i++)
#define memset(x,y) memset(x,y,sizeof(x))
#define memcpy(x,y) memcpy(x,y,sizeof(y))
#define all(x) x.begin(),x.end()
#define readc(x) scanf("%c",&x)
#define read(x) scanf("%d",&x)
#define read2(x,y) scanf("%d%d",&x,&y)
#define read3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define print(x) printf("%d\n",x)
#define lowbit(x) x&-x
#define lson(x) x<<1
#define rson(x) x<<1|1
#define pb push_back
#define mp make_pair
typedef pair<int,int> P;
typedef long long LL;
typedef long long ll;
const double eps=1e-8;
const double PI = acos(1.0);
const int INF = 0x3f3f3f3f;
const int inf = 0x3f3f3f3f;
const int MOD = 1e9+7;
const ll mod = 998244353;
const int MAXN = 1e6+7;
const int maxm = 1;
const int maxn = 100000+10;
int T;
int n,m;
int p1,p2;
int s1,s2;
int x,y;
int ans = 0;
struct node
{
  int h,m;
}a[maxn];

bool cmp(node a,node b)
{
  if(a.h == b.h)
    return a.m >= b.m;
  else
    return a.h>b.h;
}
int main()
{
    int n;
    read(n);
    node ma,mi;
    node st1,st2,st3,st4,st5,st6;
    ma.h = 6;
    ma.m = 29;
    mi.h = 19;
    mi.m = 1;
    st1.h = 6;
    st1.m = 30;
    st2.h = 19;
    st2.m = 0;
    st3.h = 10;
    st3.m = 0;
    st4.h = 10;
    st4.m = 1;
    st5.h = 16;
    st5.m = 0;
    st6.h = 16;
    st6.m = 01;

    rep(i,0,n)
    {
      node temp;
      scanf("%d:%d",&temp.h,&temp.m);
      if(cmp(st1,temp)||cmp(temp,st2))
        continue;
      if(cmp(temp,ma))
      {
        ma.h = temp.h;
        ma.m = temp.m;
      }
      if(cmp(mi,temp))
      {
        mi.h = temp.h;
        mi.m = temp.m;
      }
    }
    // printf("%d:%d\n",ma.h,ma.m);
    // printf("%d:%d\n",mi.h,mi.m);
    if(cmp(mi,st1) && cmp(st3,mi) && cmp(ma,st1) && cmp(st5,ma))
      printf("24000\n");
    else if(cmp(mi,st1) && cmp(st3,mi) && cmp(ma,st6) && cmp(st2,ma))
      printf("36000\n");
    else if(cmp(mi,st4) && cmp(st5,mi) && cmp(ma,st4) && cmp(st5,ma) )
      printf("16800\n");
    else if(cmp(mi,st4) && cmp(st2,mi) && cmp(ma,st6) && cmp(st2,ma))
      printf("24000\n");
    else
      printf("0\n");

}