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导入(ibd文件)
查看>>
Mysql工作笔记006---Mysql服务器磁盘爆满了_java.sql.SQLException: Error writing file ‘tmp/MYfXO41p‘
查看>>
MySQL工具1:mysqladmin
查看>>
mysql常用命令
查看>>
MySQL常用命令
查看>>
mysql常用命令
查看>>
MySQL常用指令集
查看>>
mysql常用操作
查看>>
MySQL常用日期格式转换函数、字符串函数、聚合函数详
查看>>
MySQL常见函数
查看>>
MySQL常见架构的应用
查看>>
MySQL常见的三种存储引擎(InnoDB、MyISAM、MEMORY)
查看>>
MySQL常见的三种存储引擎(InnoDB、MyISAM、MEMORY)
查看>>
MySQL常见约束条件
查看>>
MySQL常见错误
查看>>
MySQL常见错误分析与解决方法总结
查看>>
mysql并发死锁案例
查看>>
MySQL幻读:大家好,我是幻读,我今天又被解决了
查看>>
MySQL底层概述—1.InnoDB内存结构
查看>>
MySQL底层概述—2.InnoDB磁盘结构
查看>>