Codeforces Round #284 (Div. 2)

题目链接：http://codeforces.com/contest/499

A. Watching a movie

You have decided to watch the best moments of some movie. There are two buttons on your player:

1. Watch the current minute of the movie. By pressing this button, you watch the current minute of the movie and the player automatically proceeds to the next minute of the movie.
2. Skip exactly x minutes of the movie (x is some fixed positive integer). If the player is now at the t-th minute of the movie, then as a result of pressing this button, it proceeds to the minute (t + x).

Initially the movie is turned on in the player on the first minute, and you want to watch exactly n best moments of the movie, the i-th best moment starts at the li-th minute and ends at the ri-th minute (more formally, the i-th best moment consists of minutes: li, li + 1, ..., ri).

Determine, what is the minimum number of minutes of the movie you have to watch if you want to watch all the best moments?

题意：你正在观看电影，现在有两个按钮1和2，如果你在电影t分钟时选择按钮1，电影会进入到t+1分钟；如果你在t分钟选择按钮2，电影在跳到t+x分钟。现在给出了电影中n个你想要看的精彩连续片段，假如你此时在电影的开头（即第一分钟），求出你最少要观看电影多少分钟。

解法：题目中给出的n个片段[Li,Ri]都是按照时间顺序排好了的（Ri-1<Li），所以我们不需要排序，直接模拟这个过程即可。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<cstdlib>
 5 #include<cmath>
 6 #include<algorithm>
 7 #define inf 0x7fffffff
 8 using namespace std;
 9 const int maxn=55;
10 int main()
11 {
12     int n;
13     int l,L,R;
14     int x;
15     while (scanf("%d%d",&n,&x)!=EOF)
16     {
17         int total=0;
18         l=1;
19         for (int i=0 ;i<n ;i++)
20         {
21             scanf("%d%d",&L,&R);
22             while (l+x<=L) {l+=x;if (l+x>L) break; }
23             total += R-l+1;
24             l=R+1;
25         }
26         printf("%d\n",total);
27     }
28     return 0;
29 }

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