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 InnoDB 三大文件日志,看完秒懂
查看>>
Mysql InnoDB 数据更新导致锁表
查看>>
Mysql Innodb 锁机制
查看>>
MySQL InnoDB中意向锁的作用及原理探
查看>>
MySQL InnoDB事务隔离级别与锁机制深入解析
查看>>
Mysql InnoDB存储引擎 —— 数据页
查看>>
Mysql InnoDB存储引擎中的checkpoint技术
查看>>
Mysql InnoDB存储引擎中缓冲池Buffer Pool、Redo Log、Bin Log、Undo Log、Channge Buffer
查看>>
MySQL InnoDB引擎的锁机制详解
查看>>
Mysql INNODB引擎行锁的3种算法 Record Lock Next-Key Lock Grap Lock
查看>>
mysql InnoDB数据存储引擎 的B+树索引原理
查看>>
mysql innodb通过使用mvcc来实现可重复读
查看>>
mysql insert update 同时执行_MySQL进阶三板斧(三)看清“触发器 (Trigger)”的真实面目...
查看>>
mysql interval显示条件值_MySQL INTERVAL关键字可以使用哪些不同的单位值?
查看>>
Mysql join原理
查看>>
MySQL Join算法与调优白皮书(二)
查看>>
Mysql order by与limit混用陷阱
查看>>
Mysql order by与limit混用陷阱
查看>>
mysql order by多个字段排序
查看>>
MySQL Order By实现原理分析和Filesort优化
查看>>