PAT(Phone Bills)

2018-08-15  本文已影响8人  小幸运Q

参考:https://blog.csdn.net/xyt8023y/article/details/46010349

1. 题目有点奇葩：

为了解决on和off存储的问题，我选择使用正负号来区分on和off。

分以下四种情况：
i指向on，j指向off
++    i++  j++
--    i=j+1 j+=2
+-    i=j+1 j+=2
-+    i++  j++

我的流程：

将输入写入vector后按照字母次序排序：

// 打印的时候要按照字符大小排序(ascii码小的放前面)
bool cmpstr(customer&left,customer&right){
  // 逐个字符比较,从小到大排序,防止前面一串出现重复
  int leftlen=strlen(left.name);
  int rightlen=strlen(right.name);
  int length=min(leftlen,rightlen);
  int i,j;
  for(i=0;i<length;i++){
    if(left.name[i]==right.name[i]){

    }
    else{
      return left.name[i]<right.name[i];
    }
  }
  return leftlen<rightlen;
}

数据清理：

对每一个人的time进行排序：

  // 按照题目要求的字符串方式排序
  sort(v.begin(),v.end(),cmpstr);
  for(i=0;i<v.size();i++){
    // 对时间从小到大排序
    sort(v[i].time.begin(),v[i].time.end(),cmptime);
  }

打印每个人的数据：

边过滤边将合法的on-off插入vector中（记得取绝对值）：

void print(vector<customer>&v,int num){
  int i=0,j=1;
  float sum=0;
  // 计算总价
  vector<int>record;
  while(j<v[num].time.size()){
    if(v[num].time[i]>0&&v[num].time[j]<0){
      int start=v[num].time[i];
      int ending=v[num].time[j];
      record.push_back(start);
      record.push_back(ending);
    }
    if(v[num].time[j]>0){
      i++;
      j++;
    }
    else if(v[num].time[j]<0){
      i=j+1;
      j+=2;
    }
  }
  if(record.empty()){
    return;
  }
  printf("%s %02d\n",v[num].name,monthnow);
  for(i=0;i<record.size();i+=2){
    PrintTime(record[i]);
    PrintTime(abs(record[i+1]));
    int time=-(record[i]+record[i+1]);
    cout<<time<<" ";
    sum+=sumup(record[i],abs(record[i+1]));
  }
  printf("Total amount: $%.2f\n",sum);
}

根据开始结束时间计算收费

float sumup(int start,int ending){
  int i,j;
  int start_hours=start/60;
  int start_minutes=start%60;
  int ending_hours=ending/60;
  int ending_minutes=ending%60;
  float cost=0;
  for(i=start_hours+1;i<ending_hours;i++){
    cost+=60*costs[i%24];
  }
  cost+=costs[ending_hours%24]*(ending_minutes);
  cost+=costs[start_hours%24]*(60-start_minutes);
  cost/=100;
  printf("$%.2f\n",cost);
  return cost;
}

以上模块有两个问题，一个在计算中不提前除以100.00的话有可能会溢出。一个是算法错了，在同一个小时内的call会出现计算错误，下面这个算法计算了start到0：0：0的开销如何减去ending到0：0：0的开销得出了正确的结论。

double caculate(int t)
{
    int i,j;
    double sum=0;
    for(i=0;i<t/60;i++)
    {
        sum+=costs[i%24]*60/100.0;
    }
    sum+=costs[i%24]*(t%60)/100.0;
    return sum;
}
double sumup(int start,int ending)
{
    double cost=caculate(ending)-caculate(start);
    printf("$%.2lf\n",cost);
    return cost;
}

我的代码：

#include<bits/stdc++.h>
using namespace std;
typedef struct customer{
  vector<int>time;
  // +on -off
  string name;
  int namehash;
}customer;
map<string,customer>m;

int costs[24]={0};
int monthnow=0;
// 月份是固定的,单独存放
vector<customer>v;

// 查找hash值是否存在，不存在就返回-1
int searchmap(char name[]){
  int i,j;
  if(m.count(name)==1){
    return 1;
  }
  return -1;
}
// 根据名字hash从小到大排列，方便二叉查找
bool cmp(customer&left,customer&right){
  return left.namehash<right.namehash;
}
// 打印的时候要按照字符大小排序(ascii码小的放前面)
bool cmpstr(customer&left,customer&right){
  // 逐个字符比较,从小到大排序,防止前面一串出现重复
  int leftlen=left.name.size();
  int rightlen=right.name.size();
  int length=min(leftlen,rightlen);
  int i,j;
  for(i=0;i<length;i++){
    if(left.name[i]==right.name[i]){

    }
    else{
      return left.name[i]<right.name[i];
    }
  }
  return leftlen<rightlen;
}
bool cmptime(int a,int b){
  return abs(a)<abs(b);
}
void PrintTime(int time){
  printf("%02d:%02d:%02d ",time/(60*24),(time%(60*24))/60,(time%(60*24))%60);
}
double caculate(int t)
{
    int i,j;
    double sum=0;
    for(i=0;i<t/60;i++)
    {
        sum+=costs[i%24]*60/100.0;
    }
    sum+=costs[i%24]*(t%60)/100.0;
    return sum;
}
double sumup(int start,int ending)
{
    double cost=caculate(ending)-caculate(start);
    printf("$%.2lf\n",cost);
    return cost;
}
void print(vector<customer>&v,int num){
  int i=0,j=1;
  double sum=0;
  // 计算总价
  vector<int>record;
  while(j<v[num].time.size()){
    if(v[num].time[i]>0&&v[num].time[j]<0){
      int start=v[num].time[i];
      int ending=v[num].time[j];
      record.push_back(start);
      record.push_back(ending);
    }
    if(v[num].time[j]>0){
      i++;
      j++;
    }
    else if(v[num].time[j]<0){
      i=j+1;
      j+=2;
    }
  }
  if(record.empty()){
    return;
  }
  printf("%s %02d\n",v[num].name.c_str(),monthnow);
  for(i=0;i<record.size();i+=2){
    PrintTime(record[i]);
    PrintTime(abs(record[i+1]));
    int time=-(record[i]+record[i+1]);
    cout<<time<<" ";
    sum+=sumup(record[i],abs(record[i+1]));
  }
  printf("Total amount: $%.2f\n",sum);
}
int main(){
  int i,j,input;
  for(i=0;i<24;i++){
    scanf("%d",&input);
    costs[i]=input;
  }
  scanf("%d",&input);

  char name[50],sign[20];
  int month,day,hour,minutes,namehash;
  for(j=0;j<input;j++){
    scanf("%s %d:%d:%d:%d %s",name,&month,&day,&hour,&minutes,sign);
    monthnow=month;
    // cout<<"))))))))"<<endl;
    if(!strcmp(sign,"off-line")){
      // 匹配offline成功则返回0
      int target=searchmap(name);
      if(target==-1){
        // 没找到
        customer c;
        c.name=name;
        //c.namehash=hashname(name);
        c.time.push_back((minutes+60*hour+24*60*day)*-1);
        m[name]=c;
        //v.push_back(c);
        // sort(v.begin(),v.end(),cmpstr);
      }
      else{
        m[name].time.push_back((minutes+60*hour+24*60*day)*-1);
        // v[target].time.push_back((minutes+60*hour+24*60*day)*-1);
      }
    }
    else{
      int target=searchmap(name);
      if(target==-1){
        // 没找到
        customer c;
        c.name=name;
        c.time.push_back(minutes+60*hour+24*60*day);
        m[name]=c;
      }
      else{
        m[name].time.push_back(minutes+60*hour+24*60*day);
      }
    }
  }
  for(map<string,customer>::iterator it=m.begin();it!=m.end();it++){
    v.push_back(it->second);
  }
  // 按照题目要求的字符串方式排序
  sort(v.begin(),v.end(),cmpstr);
  for(i=0;i<v.size();i++){
    // 对时间从小到大排序
    sort(v[i].time.begin(),v[i].time.end(),cmptime);
  }
  // 筛选每个customers的bills然后打印
  for(i=0;i<v.size();i++){
    print(v,i);
  }
}

测试数据：


测试用例:
72 53 94 95 71 82 31 1 97 36 8 94 44 22 3 51 58 72 27 57 7 39 55 80 38 
CS07 07:11:04:05 off-line 
CS05 07:15:19:37 on-line 
CS09 07:26:10:05 on-line 
CS00 07:29:08:20 off-line 
CS00 07:27:16:05 on-line 
CS06 07:25:13:50 off-line 
CS07 07:17:08:03 on-line 
CS04 07:25:02:50 on-line 
CS02 07:05:09:47 on-line 
CS03 07:23:18:20 on-line 
CS05 07:30:05:51 on-line 
CS06 07:08:11:57 off-line 
CS05 07:13:18:12 on-line 
CS07 07:21:07:52 off-line 
CS08 07:14:12:34 off-line 
CS04 07:21:18:49 off-line 
CS00 07:31:11:35 on-line 
CS09 07:03:10:58 on-line 
CS08 07:17:01:46 off-line 
CS06 07:02:02:30 off-line 
CS07 07:16:08:51 on-line 
CS02 07:29:17:32 off-line 
CS03 07:14:19:43 on-line 
CS05 07:12:06:43 off-line 
CS09 07:25:04:45 off-line 
CS04 07:23:15:19 on-line 
CS08 07:03:10:42 on-line 
CS03 07:01:12:02 off-line 
CS07 07:30:07:23 off-line 
CS03 07:28:14:33 off-line 
CS00 07:08:18:33 off-line 
CS08 07:13:13:44 off-line 
CS02 07:30:13:47 on-line 
CS06 07:14:21:54 on-line 
CS00 07:09:21:00 on-line 
CS02 07:24:12:24 off-line 
CS09 07:24:19:24 off-line 
CS03 07:16:02:33 on-line 
对应输出应该为:
CS00 07 27:16:05 29:08:20 2415 $1302.30 Total amount: $1302.30 
CS02 07 05:09:47 24:12:24 27517 $14315.04 Total amount: $14315.04 
CS03 07 23:18:20 28:14:33 6973 $3632.19 Total amount: $3632.19 
CS06 07 14:21:54 25:13:50 15356 $8055.14 Total amount: $8055.14 
CS07 07 17:08:03 21:07:52 5749 $2994.61 Total amount: $2994.61 
CS08 07 03:10:42 13:13:44 14582 $7587.92 Total amount: $7587.92 
CS09 07 03:10:58 24:19:24 30746 $15973.84 Total amount: $15973.84 
你的输出为:
CS00 07 27:16:05 29:08:20 2415 $1302.30 Total amount: $1302.30 
CS02 07 05:09:47 24:12:24 27517 $14315.05
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