PAT (Advanced Level) Practice1016 Phone Bills

最近初学c++来练习算法题，将pta 的题目重新用c++写一遍，顺便熟悉c++常用stl库及api

电话账单问题（模拟问题）

//	代码
#include
#include
#include
#include
#include
#include

using namespace std;
const int N = 1010, M = 31 * 1440 + 10;
double cost[M];




struct Records{
    
    int minutes;
    string state;
    string format_time;
    
    bool operator< (const Records& t) const
    {
        return minutes < t.minutes;
    }
};

map> persons;

int main()
{
    int time[24];
    for(int i = 0; i < 24; i++)cin >> time[i];
    int n;
    for(int i = 1; i < M; i++) cost[i] = cost[i - 1] + time[(i - 1) % 1440 / 60] * 0.01;        //前缀和
    cin >> n;
    char name[30], state[30], format_time[30];
    int month, day, hour, minute;
    for(int i = 0; i < n ; i++)
    {
        scanf("%s %d:%d:%d:%d %s", name, &month, &day, &hour, &minute, state);
        
        sprintf(format_time, "%02d:%02d:%02d", day, hour, minute);
        int minutes = (day - 1) * 1440 + hour * 60 + minute;
        persons[name].push_back({minutes, state, format_time});
    }
    
    for(auto& person : persons) //通过直接调用实参，能减少创建形参的时间
    {
        string name = person.first;
        auto records = person.second;
        sort(records.begin(), records.end());
        double total = 0;
        for(int i = 0 ; i + 1 < records.size(); i++)
        {
            auto a = records[i], b = records[i + 1];
            if(a.state == "on-line" && b .state == "off-line")
            {
                if(!total)printf("%s %02dn", name.c_str(), month);
                cout << a.format_time << ' ' << b.format_time << ' ' << b.minutes - a.minutes << ' ';
                double money = cost[b.minutes] - cost[a.minutes];
                printf("$%.2lfn", money);
                total += money;
            }
            
        }
        
        if(total)
        {
            printf("Total amount: $%.2lfn", total);
        }
    }
    return 0;
}

前缀和：适用于连续区间之间的差值问题

const int N = 1010, M = 31 * 1440 + 10;
double cost[M];
for(int i = 1; i < M; i++)
    cost[i] = cost[i - 1] + time[(i - 1) % 1440 / 60] * 0.01;        //前缀和 
							//	i-1   : 分钟将会向上取整 （6:59 ~ 7 : 00 还是算在6点的价格）
							//	%1440 : 将分钟限制在一天内
							//	/60   : 将分钟划分出特定的小时段
//	定义每一分钟的和 （通过前一分钟的和与当前分钟的话费相加实现）

vector（可变长数组，有点类似Java中的ArrayList）

#include		//引入头文件
struct Recrods{
    int minutes;
    string state;
    string format_time;
    
    // 制定构造器自动排序规则
    bool opetator< (const Records& t) const
    {
        return minutes < t.minutes;
    }
}
int main()
{
    map> persons;
    for(auto person: persons )
    {
        auto records = person.second;
        
        // 通过结构体中的构造器进行自定义排序
        sort(records.begin(), records.end());
    }
    
}

scanf 、sprintf

#include
#include
#include
 char name[30], state[30], format_time[30];
    int month, day, hour, minute;
    scanf("%s %d:%d:%d:%d %s", name, &month, &day, &hour, &minute, state);
	//	scanf能根据特定的输入导入要求的数据


	sprintf(format_time, "%02d:%02d:%02d", day, hour, minute);
	//	sprintf能将特定的输入导出到元素中

PAT (Advanced Level) Practice1016 Phone Bills

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