【codeforces #10 】

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

题意：电脑正常运行的时候消耗的功率是p1每分钟，t1时间后如果不点击鼠标或键盘，电脑进入屏幕保护状态，此时的功率消耗p2每分钟，保持该状态t2时间后进入休眠状态，此时的电脑功率消耗为p3每分钟。给出n个时间段，时间段内连续点击鼠标或键盘。求最后消耗的总功率。

解题：就一道简单的数学题；直接上代码：

code：

#include <iostream>#include <algorithm>#include <cstdio>#include <string.h>using namespace std;struct node{    int l,r;}a[102];int main(){    int n,p1,p2,p3,t1,t2;    while(scanf("%d%d%d%d%d%d",&n,&p1,&p2,&p3,&t1,&t2)!=EOF)    {       int tp1=0,tp2=0,tp3=0;       for(int i=0;i<n;i++)       {           scanf("%d%d",&a[i].l,&a[i].r);       }       if(n==1)        printf("%d\n",(a[0].r-a[0].l)*p1);       else       {            tp1=a[0].r-a[0].l;           for(int i=1;i<n;i++)           {               tp1+=(a[i].r-a[i].l);               if((a[i].l-a[i-1].r)<=t1)                tp1+=(a[i].l-a[i-1].r);               else if((a[i].l-a[i-1].r)<=(t1+t2))               {                   tp1+=t1;                   tp2+=(a[i].l-a[i-1].r-t1);               }               else               {                   tp1+=t1;                   tp2+=t2;                   tp3+=(a[i].l-a[i-1].r-t1-t2);               }           }           printf("%d\n",tp1*p1+tp2*p2+tp3*p3);       }    }    return 0;}