博客
关于我
CodeForces - 10A_模拟
阅读量:136 次
发布时间:2019-02-28

本文共 2279 字,大约阅读时间需要 7 分钟。

Tom is interested in power consumption of his favourite laptop. His laptop has three modes. In normal mode laptop consumes P1 watt per minute. T1 minutes after Tom moved the mouse or touched the keyboard for the last time, a screensaver starts and power consumption changes to P2 watt per minute. Finally, after T2 minutes from the start of the screensaver, laptop switches to the “sleep” mode and consumes P3 watt per minute. If Tom moves the mouse or touches the keyboard when the laptop is in the second or in the third mode, it switches to the first (normal) mode. Tom’s work with the laptop can be divided into n time periods [l1, r1], [l2, r2], …, [ln, rn]. During each interval Tom continuously moves the mouse and presses buttons on the keyboard. Between the periods Tom stays away from the laptop. Find out the total amount of power consumed by the laptop during the period [l1, rn].

Input
The first line contains 6 integer numbers n, P1, P2, P3, T1, T2 (1 ≤ n ≤ 100, 0 ≤ P1, P2, P3 ≤ 100, 1 ≤ T1, T2 ≤ 60). The following n lines contain description of Tom’s work. Each i-th of these lines contains two space-separated integers li and ri (0 ≤ li < ri ≤ 1440, ri < li + 1 for i < n), which stand for the start and the end of the i-th period of work.
Output

Output the answer to the problem.

Examples

Input

1 3 2 1 5 100 10

Output

30

Input

2 8 4 2 5 1020 3050 100

Output

570

题目大意:一台电脑有三种工作状态,每个工作状态有不同的耗电功率,求耗电值。


这题挺考察分类细节的,一个地方错了就过不了。

inline int f(int x, int l, int r){       return x * (r - l);}int main(){       int n, p1, p2, p3, t1, t2;    cin >> n >> p1 >> p2 >> p3 >> t1 >> t2;    int ans = 0;    int last = -1;    while (n--)    {           int a, b;        cin >> a >> b;        ans += f(p1, a, b);        if (last != -1)            if (a - last <= t1)            {                   ans += f(p1, last, a);            }            else            {                   ans += f(p1, last, last + t1);                if (a - last - t1 <= t2)                {                       ans += f(p2, last + t1, a);                }                else                {                       ans += f(p2, last + t1, last + t1 + t2);                    ans += f(p3, last + t1 + t2, a);                }            }        last = b;    }    cout << ans << endl;    return 0;}

转载地址:http://jeod.baihongyu.com/

你可能感兴趣的文章
MySQL中使用IN()查询到底走不走索引?
查看>>
Mysql中使用存储过程插入decimal和时间数据递增的模拟数据
查看>>
MySql中关于geometry类型的数据_空的时候如何插入处理_需用null_空字符串插入会报错_Cannot get geometry object from dat---MySql工作笔记003
查看>>
mysql中出现Incorrect DECIMAL value: '0' for column '' at row -1错误解决方案
查看>>
mysql中出现Unit mysql.service could not be found 的解决方法
查看>>
mysql中出现update-alternatives: 错误: 候选项路径 /etc/mysql/mysql.cnf 不存在 dpkg: 处理软件包 mysql-server-8.0的解决方法(全)
查看>>
Mysql中各类锁的机制图文详细解析(全)
查看>>
MySQL中地理位置数据扩展geometry的使用心得
查看>>
Mysql中存储引擎简介、修改、查询、选择
查看>>
Mysql中存储过程、存储函数、自定义函数、变量、流程控制语句、光标/游标、定义条件和处理程序的使用示例
查看>>
mysql中实现rownum,对结果进行排序
查看>>
mysql中对于数据库的基本操作
查看>>
Mysql中常用函数的使用示例
查看>>
MySql中怎样使用case-when实现判断查询结果返回
查看>>
Mysql中怎样使用update更新某列的数据减去指定值
查看>>
Mysql中怎样设置指定ip远程访问连接
查看>>
mysql中数据表的基本操作很难嘛,由这个实验来带你从头走一遍
查看>>
Mysql中文乱码问题完美解决方案
查看>>
mysql中的 +号 和 CONCAT(str1,str2,...)
查看>>
Mysql中的 IFNULL 函数的详解
查看>>