Codeforces 10A-Power Consumption Calculation(模拟)

A. Power Consumption Calculation

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 nlines 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 test(s)

input

1 3 2 1 5 10
0 10

output

30

input

2 8 4 2 5 10
20 30
50 100

output

570

 题意：电脑有三种模式，正常模式每分钟耗电p1。假设没有使用电脑t1分钟后变成另外一种模式每分钟耗电p2，假设还是没有使用电脑t2分钟后变成第三种模式每分钟耗电p3。给定n个区间，每个区间是正常模式，每个区间的间隔是没有使用。问总的耗电是多少

思路：直接模拟就可以。

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <string>
#include <vector>
using namespace std;
#define LL long long
int main()
{
	int n,p1,p2,p3,t1,t2,l[110],r[110];
	while(cin>>n>>p1>>p2>>p3>>t1>>t2)
	{
		int ans=0;
		for(int i=0;i<n;i++)
		{
			cin>>l[i]>>r[i];
			ans+=(r[i]-l[i])*p1;
			if(i>0)
			{
				if(l[i]-r[i-1]>t1)
				{
					ans+=t1*p1;
					int t=l[i]-r[i-1]-t1;
					if(t>t2)
					{
						ans+=t2*p2;
						ans+=(t-t2)*p3;
					}
					else
						ans+=t*p2;
				}
				else
					ans+=(l[i]-r[i-1])*p1;
			}
		}
		printf("%d\n",ans);
	}
	return 0;
}