Codeforces Round #271 (Div. 2)-B. Worms
http://codeforces.com/problemset/problem/474/B
1 second
256 megabytes
standard input
standard output
It is lunch time for Mole. His friend, Marmot, prepared him a nice game for lunch.
Marmot brought Mole n ordered piles of worms such that i-th pile contains ai worms. He labeled all these worms with consecutive integers: worms in first pile are labeled with numbers 1 to a1, worms in second pile are labeled with numbers a1 + 1 to a1 + a2 and so on. See the example for a better understanding.
Mole can't eat all the worms (Marmot brought a lot) and, as we all know, Mole is blind, so Marmot tells him the labels of the best juicy worms. Marmot will only give Mole a worm if Mole says correctly in which pile this worm is contained.
Poor Mole asks for your help. For all juicy worms said by Marmot, tell Mole the correct answers.
The first line contains a single integer n (1 ≤ n ≤ 105), the number of piles.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 103, a1 + a2 + ... + an ≤ 106), where ai is the number of worms in the i-th pile.
The third line contains single integer m (1 ≤ m ≤ 105), the number of juicy worms said by Marmot.
The fourth line contains m integers q1, q2, ..., qm (1 ≤ qi ≤ a1 + a2 + ... + an), the labels of the juicy worms.
Print m lines to the standard output. The i-th line should contain an integer, representing the number of the pile where the worm labeled with the number qi is.
5
2 7 3 4 9
3
1 25 11
1
5
3
For the sample input:
- The worms with labels from [1, 2] are in the first pile.
- The worms with labels from [3, 9] are in the second pile.
- The worms with labels from [10, 12] are in the third pile.
- The worms with labels from [13, 16] are in the fourth pile.
- The worms with labels from [17, 25] are in the fifth pile.
解题思路:a1, a2, a3...an个worms分别在第1, 2, 3....n个pile上,问你q1, q2, q3...qm条worms在第几个pile上,分别算出到第1, 第2, 第3。。。 第n个pile总共有几条worms,然后二分即可
1 #include <stdio.h>
2
3 int p[];
4
5 int bin_search(int num, int l, int r){
6 int mid;
7 while(l <= r){
8 mid = (l + r) >> ;
9 if(num > p[mid-] && num <= p[mid]){
break;
}
if(p[mid] > num){
r = mid -;
}
else if(p[mid] < num){
l = mid + ;
}
}
return mid;
}
int main(){
int n, a, m, q, i;
while(scanf("%d", &n) != EOF){
scanf("%d", &p[]);
for(i = ; i < n; i++){
scanf("%d", &a);
p[i + ] = p[i] + a;
}
scanf("%d", &m);
while(m--){
scanf("%d", &q);
printf("%d\n", bin_search(q, , n));
}
}
return ;
37 }
Codeforces Round #271 (Div. 2)-B. Worms的更多相关文章
- Codeforces Round #271 (Div. 2)题解【ABCDEF】
Codeforces Round #271 (Div. 2) A - Keyboard 题意 给你一个字符串,问你这个字符串在键盘的位置往左边挪一位,或者往右边挪一位字符,这个字符串是什么样子 题解 ...
- Codeforces Round #271 (Div. 2) 解题报告
题目地址:http://codeforces.com/contest/474 A题:Keyboard 模拟水题. 代码例如以下: #include <iostream> #include ...
- Codeforces Round #271 (Div. 2)
A. Keyboard 题意:一个人打字,可能会左偏一位,可能会右偏一位,给出一串字符,求它本来的串 和紫书的破损的键盘一样 #include<iostream> #include< ...
- Codeforces Round #271 (Div. 2) F. Ant colony (RMQ or 线段树)
题目链接:http://codeforces.com/contest/474/problem/F 题意简而言之就是问你区间l到r之间有多少个数能整除区间内除了这个数的其他的数,然后区间长度减去数的个数 ...
- Codeforces Round #271 (Div. 2) D. Flowers (递推)
题目链接:http://codeforces.com/problemset/problem/474/D 用RW组成字符串,要求w的个数要k个连续出现,R任意,问字符串长度为[a, b]时,字符串的种类 ...
- Codeforces Round #271 (Div. 2) E题 Pillars(线段树维护DP)
题目地址:http://codeforces.com/contest/474/problem/E 第一次遇到这样的用线段树来维护DP的题目.ASC中也遇到过,当时也非常自然的想到了线段树维护DP,可是 ...
- Codeforces Round #271 (Div. 2) F题 Ant colony(线段树)
题目地址:http://codeforces.com/contest/474/problem/F 由题意可知,最后能够留下来的一定是区间最小gcd. 那就转化成了该区间内与区间最小gcd数相等的个数. ...
- Codeforces Round #271 (Div. 2)-A. Keyboard
http://codeforces.com/problemset/problem/474/A A. Keyboard time limit per test 2 seconds memory limi ...
- Codeforces Round #271 (Div. 2) F ,E, D, C, B, A
前言:最近被线段树+简单递推DP虐的体无完肤!真是弱! A:简单题,照着模拟就可以,题目还特意说不用处理边界 B:二分查找即可,用lower_lound()函数很好用 #include<stri ...
随机推荐
- HTML基本标签元素
HTML: 超文本标记语言(HyperText Mark-up Language ) 1.作用:写网页结构 2.HTML不区分大小写,建议小写 3.文件后缀 .html 或者 .ht ...
- 使用Bootstrap实现表格列的显示与隐藏
来人,上效果图 走官方通道 1.引入样式文件(去github下载样式文件) <!--插件开--> <link rel="stylesheet" href=&quo ...
- 什么是socket ??
socket起源于Unix, 而Unix/Linux基本哲学之一就是"一切皆文件", 都可以用"打开open -> 读写write/read -> 关闭clo ...
- PJzhang:尽快发现并下架那些侵犯公司权利的假冒APP
猫宁!!! 参考链接:https://www.freebuf.com/articles/paper/203358.html http://www.cac.gov.cn/2019-01/25/c_112 ...
- 在接口的实现类里使用@Override注解报错
问题分析 @Override注解用来检测子类对父类或接口的方法的重写是否正确,但有一次我在Eclipse里对接口的实现类里使用@Override注解却报错,不过在父类的子类里使用该注解却是正常的. 百 ...
- oracle 数据库 存储过程
destinct(去除重复行), to_char(转换自符串), to_date(日期格式函数), substr(截取自符串), lower(将字母转换为小写), upper(将字母转换为小写), t ...
- [已读]移动web手册
冲着作者是PPK才买的,双色,180页,略贵.但是这本书的内容,我并不是很满意.4.5.6三章算是干货,尤其是第四章,值得一读.其他的内容,忽略就好了.
- mac 终端查看端口命令
查看 端口所在线程 lsof -i:8080 mac-abeen:spider abeen$ lsof -i:8080 COMMAND PID USER FD TYPE DEVICE SIZE/OFF ...
- python_21(Django中间件)
第1章 中间件 1.1 介绍 1.2 种类 1.3 自定义中间件 1.4 process_request 1.4.1 注册中间件 1.5 process_response 1.6 process_vi ...
- iis日志存放位置 及 查看方法
IIS:控制面板--管理工具--internet信息服务 网站的IIS日志是在空间里面看的.要登陆到空间里面的一个IIS日志里面看.IIS日志一般都很大的.看会有点.. 一.应用程序日志.安全日志.系 ...