Codeforces Beta Round #10 A. Power Consumption Calculation 水题

A. Power Consumption Calculation

题目连接：

http://www.codeforces.com/contest/10/problem/A

Description

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.

Sample Input

1 3 2 1 5 10

0 10

Sample Output

30

Hint

题意

有一台电脑，在活跃状态下，每秒钟耗能p1，在没人碰之后的t1秒后，进入休息状态，每秒钟耗能p2，如果再有t3秒没人碰的话，就会进入睡眠状态，每秒钟耗能p3

给你这个人碰电脑的区间时间，问你这台电脑的耗能是多少

题解：

数据范围太小，直接暴力就好了……

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 105;
int n,p1,p2,p3,t1,t2;
pair<int,int> Q[maxn];
int p[3000];
int main()
{
    scanf("%d%d%d%d%d%d",&n,&p1,&p2,&p3,&t1,&t2);
    for(int i=1;i<=n;i++)scanf("%d%d",&Q[i].first,&Q[i].second);
    for(int i=1;i<=n;i++)for(int j=Q[i].first;j<=Q[i].second;j++)
        p[j]=1;
    long long ans = 0;
    int type=0,now=0;
    for(int i=Q[1].first;i<Q[n].second;i++)
    {
        if(p[i]==1)now=0,ans+=p1;
        else if(now>=t1+t2)ans+=p3;
        else if(now>=t1)ans+=p2;
        else ans+=p1;
        now++;
    }
    cout<<ans<<endl;
}