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当前列的值等于上一行的值累加前一列的值
查看>>
MySQL当查询的时候有多个结果,但需要返回一条的情况用GROUP_CONCAT拼接
查看>>
MySQL必知必会(组合Where子句,Not和In操作符)
查看>>
MySQL必知必会总结笔记
查看>>
MySQL快速入门
查看>>
MySQL快速入门——库的操作
查看>>
mysql快速复制一张表的内容,并添加新内容到另一张表中
查看>>
mysql快速查询表的结构和注释,字段等信息
查看>>
mysql怎么删除临时表里的数据_MySQL中关于临时表的一些基本使用方法
查看>>
mysql性能优化
查看>>
mysql性能优化学习笔记-存储引擎
查看>>
MySQL性能优化必备25条
查看>>
Mysql性能优化(1):SQL的执行过程
查看>>
Mysql性能优化(2):数据库索引
查看>>
Mysql性能优化(3):分析执行计划
查看>>
Mysql性能优化(4):优化的注意事项
查看>>
Mysql性能优化(5):主从同步原理与实现
查看>>
Mysql性能优化(6):读写分离
查看>>
MySQL性能优化(八)--
查看>>
MySQL性能测试及调优中的死锁处理方法
查看>>