【水分段函数】#10 A.Power Consumption Calculation

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 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 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

这道题叫“功耗计算”，大概意思呢是告诉你这么件事——

当你afk时，电脑的耗电大概是这么回事：

然而期间只要你动了一下电脑，立马从零开始计时。问你这个Tom在这么几个阶段里耗了多少电。

嘛~ 大概意思就是个水题啦~^_^  函数分定义域计算的问题似乎是初中……还是小学来着的……那个什么出租车开车算路费的问题吧~

Python：

# input
list = raw_input().split()
n,p1,p2,t1,t2,t3 = map(int , list)

# solve
ans = 0
pre = -1

while n > 0:
    n -= 1
    list = raw_input().split()
    start,end = map(int , list)
    ans += (end-start)*p1
    
    if pre != -1:
       x = start-pre
       if x > t1:
          ans += t1*p1
          x -= t1
          if x > t2:
             ans += t2*p2
             x -= t2
             ans += x*p3
          else:
             ans += x*p2
       else:
          ans += x*p1 
    pre = end

print ans

C++：

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;


int main()
{
    int n,p1,p2,p3,t1,t2 ;
    int l,r,l2,r2;
	scanf("%d%d%d%d%d%d",&n,&p1,&p2,&p3,&t1,&t2);
	scanf("%d%d", &l, &r);
	int sum = (r - l) * p1;
	while(--n)
	{
		scanf("%d%d",&l2,&r2);
        int mid = l2 - r ;
	    int p1pow = t1 * p1, p2pow = t2 * p2;
	    if (mid <= t1)
			sum += mid * p1;
	    else
	    {
		 sum += p1pow;
		 mid -= t1;
		 if (mid <= t2)
			sum += mid * p2;
		 else
			sum += p2pow + (mid - t2) * p3;
        }
	    sum += (r2 - l2) * p1;
	    l = l2;
	    r = r2;
	}
	 printf("%d",sum);	
	 return 0;
}