upc组队赛3 Congestion Charging Zon【模拟】

一时失言乱红尘 2022-01-06 07:29 234阅读 0赞

#  #

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.  
[点此查看图片][Link 1]  
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");
     
     
    }

转载于:https://www.cnblogs.com/llke/p/10799949.html

[Link 1]: https://www.cnblogs.com/llke/gallery/image/240886.html