0
点赞
收藏
分享

微信扫一扫

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



A. Power Consumption Calculation



http://codeforces.com/problemset/problem/10/A



time limit per test



memory limit per test



input



output


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



nP1P2P3T1T2 (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



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;
}




举报

相关推荐

0 条评论