cf499A-Watching a movie
http://codeforces.com/problemset/problem/499/A
You have decided to watch the best moments of some movie. There are two buttons on your player:
- 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.
- 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?
The first line contains two space-separated integers n, x (1 ≤ n ≤ 50, 1 ≤ x ≤ 105) — the number of the best moments of the movie and the value of x for the second button.
The following n lines contain the descriptions of the best moments of the movie, the i-th line of the description contains two integers separated by a space li, ri (1 ≤ li ≤ ri ≤ 105).
It is guaranteed that for all integers i from 2 to n the following condition holds: ri - 1 < li.
Output a single number — the answer to the problem.
2 3
5 6
10 12
6
1 1
1 100000
100000
In the first sample, the player was initially standing on the first minute. As the minutes from the 1-st to the 4-th one don't contain interesting moments, we press the second button. Now we can not press the second button and skip 3 more minutes, because some of them contain interesting moments. Therefore, we watch the movie from the 4-th to the 6-th minute, after that the current time is 7. Similarly, we again skip 3 minutes and then watch from the 10-th to the 12-th minute of the movie. In total, we watch 6 minutes of the movie.
In the second sample, the movie is very interesting, so you'll have to watch all 100000 minutes of the movie.
代码:
#include <fstream>
#include <iostream> using namespace std; int main(){
//freopen("D:\\input.in","r",stdin);
//freopen("D:\\output.out","w",stdout);
int n,x,l,r,t=,ans=;
scanf("%d%d",&n,&x);
for(int i=;i<=n;i++){
scanf("%d%d",&l,&r);
ans+=(l-t)%x+r-l+;
t=r+;
}
printf("%d\n",ans);
return ;
}
cf499A-Watching a movie的更多相关文章
- 题解 CF499A 【Watching a movie】
题意 看一部电影,从第一分钟开始看,你每次可以按下快进键,快进x分钟,电影有n个精彩片段,问看到所有精彩片段至少需要看多少分钟电影.(数据保证能看到所有精彩片段) 我们可以发现答案其实 就是间隔时间% ...
- Codeforces Round #219 (Div. 1) C. Watching Fireworks is Fun
C. Watching Fireworks is Fun time limit per test 4 seconds memory limit per test 256 megabytes input ...
- Codeforces Round #219 (Div. 2) E. Watching Fireworks is Fun
http://codeforces.com/contest/373/problem/E E. Watching Fireworks is Fun time limit per test 4 secon ...
- 6656 Watching the Kangaroo
6656 Watching the KangarooDay by day number of Kangaroos is decreasing just liketiger, whale or lion ...
- Uva 10339 - Watching Watches【数论,暴力】
题目链接:10339 - Watching Watches 题意:两个时钟,一个每天慢a秒,一个每天慢b秒,问两钟重新相遇的时刻 1圈有12 * 60 * 60秒,然后1圈 / abs(a - b), ...
- Error watching file for changes: EMFILE
运行reactnative项目时在编译过程中报错 Error watching file for changes: EMFILE 故障原因: 是升级后watchman不可用了,需要重装watchman ...
- 每日英语:In Digital Era, What Does 'Watching TV' Even Mean?
We spend a full five hours and 16 minutes a day in front of a screen, and that's without even turnin ...
- CF372C Watching Fireworks is Fun(单调队列优化DP)
A festival will be held in a town's main street. There are n sections in the main street. The sectio ...
- Django报错Watching for file changes with StatReloader
Django项目运行时出现:Watching for file changes with StatReloader错误 原因:环境里的django或者python的版本有问题 解决方案:升级或者降级D ...
- F - Watching Fireworks is Fun
C. Watching Fireworks is Fun 题目大意: 一个城镇有n个区域,从左到右1-n,每个区域之间距离1个单位距离.节日中有m个烟火要放,给定放的地点a[ i ].时间t[ i ] ...
随机推荐
- chrome安装vue-devtools
安装方法1: 需正常打开chrome商店,搜索vuejs devtools 安装.chrome://extensions/ 开发者工具-扩展程序下启用: 方法2: github下载插件,npm包安装依 ...
- Serf 了解
Introduction to Serf Welcome to the intro guide to Serf! This guide will show you what Serf is, expl ...
- Renesas CAT CONFIG
CAT CELL "sf_cellular_api.h" typedef enum e_sf_cellular_at_cmd_index { SF_CELLULAR_AT_CMD_ ...
- nginx限制请求之一:(ngx_http_limit_conn_module)模块
相关文章: <高可用服务设计之二:Rate limiting 限流与降级> <nginx限制请求之一:(ngx_http_limit_conn_module)模块> <n ...
- 杂项:Hadoop
ylbtech-杂项:Hadoop Hadoop是一个由Apache基金会所开发的分布式系统基础架构.用户可以在不了解分布式底层细节的情况下,开发分布式程序.充分利用集群的威力进行高速运算和存储.Ha ...
- 常见的sql server 链接问题------持续更新
问题1:超时时间已到.超时时间已到,但是尚未从池中获取连接.出现这种情况可能是因为所有池连接均在使用,并且达到了最大池大小 再查询窗口输入exec sp_who2进行查询链接消耗资源 可能出现的情况是 ...
- 013:Rank、视图、触发器、MySQL内建函数
一. Rank 给出不同的用户的分数,然后根据分数计算排名 (gcdb@localhost) 09:34:47 [mytest]> create table t_rank(id int,scor ...
- Hibernate保存Blob和Clob类型的数据
虽然非常不建议在数据库中保存Blob和Clob类型的数据,但真的要有这样的需求呢?这里记录一下使用Hibernate如何向数据库中保存Blob和Clob数据. Oracle和MySql在Blob类型上 ...
- Mybatis动态构建Sql(无实体类)
MyBatis的动态SQL是基于OGNL表达式的,它可以帮助我们方便的在SQL语句中实现某些逻辑. 例如,sql语句where条件中,需要一些安全判断,例如按某一条件查询时如果传入的参数是空,此时查询 ...
- 100.64.0.0/10运营商级(Carrier-grade)NAT保留IP地址
在一次跟踪路由的网络操作时发现自己路由器下一跳路由节点的IP地址比较奇怪,是100.64.0.1.好奇促使我查询了这个IP地址的归属,结果是保留地址,到这里觉得比较奇怪了,按照常理以IPv4为例保留的 ...