Codeforces Beta Round #10 / 10A Power Consumption Calculation (模拟)

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本文介绍了一个关于笔记本电脑能耗计算的问题，根据三种工作模式及其切换条件，通过输入的时间段计算总能耗。介绍了输入输出格式及样例。

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], ..., [l_n, r_n]. 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, r_n].

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 l_i and r_i (0 ≤ l_i < r_i ≤ 1440, r_i < l_i + 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

注意t2是接在t1后面算的。

完整代码：

/*30ms,0KB*/

#include<cstdio>

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