Codeforces Round #394 (Div. 2) D. Dasha and Very Difficult Problem 贪心
D. Dasha and Very Difficult Problem
题目连接:
http://codeforces.com/contest/761/problem/D
Description
Dasha logged into the system and began to solve problems. One of them is as follows:
Given two sequences a and b of length n each you need to write a sequence c of length n, the i-th element of which is calculated as follows: ci = bi - ai.
About sequences a and b we know that their elements are in the range from l to r. More formally, elements satisfy the following conditions: l ≤ ai ≤ r and l ≤ bi ≤ r. About sequence c we know that all its elements are distinct.
Dasha wrote a solution to that problem quickly, but checking her work on the standard test was not so easy. Due to an error in the test system only the sequence a and the compressed sequence of the sequence c were known from that test.
Let's give the definition to a compressed sequence. A compressed sequence of sequence c of length n is a sequence p of length n, so that pi equals to the number of integers which are less than or equal to ci in the sequence c. For example, for the sequence c = [250, 200, 300, 100, 50] the compressed sequence will be p = [4, 3, 5, 2, 1]. Pay attention that in c all integers are distinct. Consequently, the compressed sequence contains all integers from 1 to n inclusively.
Help Dasha to find any sequence b for which the calculated compressed sequence of sequence c is correct.
Input
The first line contains three integers n, l, r (1 ≤ n ≤ 105, 1 ≤ l ≤ r ≤ 109) — the length of the sequence and boundaries of the segment where the elements of sequences a and b are.
The next line contains n integers a1, a2, ..., an (l ≤ ai ≤ r) — the elements of the sequence a.
The next line contains n distinct integers p1, p2, ..., pn (1 ≤ pi ≤ n) — the compressed sequence of the sequence c.
Output
If there is no the suitable sequence b, then in the only line print "-1".
Otherwise, in the only line print n integers — the elements of any suitable sequence b.
Sample Input
5 1 5
1 1 1 1 1
3 1 5 4 2
Sample Output
3 1 5 4 2
Hint
题意
c[i]=b[i]-a[i],现在给你a[i]和c[i]的相对大小,问你可不可能出现b[i]满足条件,如果有的话,输出。
题解:
贪心,最小的显然要最小,次小的在比最小大的基础上最小就好了。
对于每一个数都二分一下就完了。
代码
#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e5+7;
long long l,r,a[maxn],b[maxn],c[maxn];
priority_queue<pair<int,int> >S;
int n,op[maxn];
int main()
{
scanf("%d%lld%lld",&n,&l,&r);
for(int i=0;i<n;i++){
scanf("%lld",&a[i]);
}
for(int i=0;i<n;i++)
scanf("%d",&op[i]),S.push(make_pair(op[i],i));
long long last = -1e16;
while(!S.empty()){
long long now = S.top().second;
S.pop();
long long L=l,R=r,Ans=r+1;
while(L<=R){
long long mid=(L+R)/2;
if(a[now]-mid>last){
L=mid+1,Ans=mid;
}else
R=mid-1;
}
if(Ans==r+1){
cout<<"-1"<<endl;
return 0;
}
b[now]=Ans;
last=a[now]-Ans;
}
for(int i=0;i<n;i++)
cout<<b[i]<<" ";
cout<<endl;
}
Codeforces Round #394 (Div. 2) D. Dasha and Very Difficult Problem 贪心的更多相关文章
- Codeforces Round #394 (Div. 2) D. Dasha and Very Difficult Problem —— 贪心
题目链接:http://codeforces.com/contest/761/problem/D D. Dasha and Very Difficult Problem time limit per ...
- Codeforces Round #394 (Div. 2) D. Dasha and Very Difficult Problem
D. Dasha and Very Difficult Problem time limit per test:2 seconds memory limit per test:256 megabyte ...
- Codeforces Round #394 (Div. 2) E. Dasha and Puzzle(分形)
E. Dasha and Puzzle time limit per test 2 seconds memory limit per test 256 megabytes input standard ...
- Codeforces Round #394 (Div. 2) E. Dasha and Puzzle 构造
E. Dasha and Puzzle 题目连接: http://codeforces.com/contest/761/problem/E Description Dasha decided to h ...
- Codeforces Round #394 (Div. 2) C. Dasha and Password 暴力
C. Dasha and Password 题目连接: http://codeforces.com/contest/761/problem/C Description After overcoming ...
- Codeforces Round #394 (Div. 2) B. Dasha and friends 暴力
B. Dasha and friends 题目连接: http://codeforces.com/contest/761/problem/B Description Running with barr ...
- Codeforces Round #394 (Div. 2) A. Dasha and Stairs 水题
A. Dasha and Stairs 题目连接: http://codeforces.com/contest/761/problem/A Description On her way to prog ...
- Codeforces Round #394 (Div. 2) C. Dasha and Password —— 枚举
题目链接:http://codeforces.com/problemset/problem/761/C C. Dasha and Password time limit per test 2 seco ...
- Codeforces Round #394 (Div. 2) B. Dasha and friends —— 暴力 or 最小表示法
题目链接:http://codeforces.com/contest/761/problem/B B. Dasha and friends time limit per test 2 seconds ...
随机推荐
- Java获取时间,将当前时间减一年,减一天,减一个月
在Java中操作时间的时候,需要计算某段时间开始到结束的区间日期,常用的时间工具 Date date = new Date();//获取当前时间 Calendar calendar = Calenda ...
- scale.fix.js
无意间在一个网站上看到的,本来是对另一个效果感兴趣的,结果看到这个放开来的js就读了一下. var metas = document.getElementsByTagName('meta'); var ...
- linq.js - LINQ for JavaScript
var jsonArray = [ { "user": { "id": 100, "screen_name": "d_linq&q ...
- html5 canvas从圆开始
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/ ...
- 关于Asp.Net中的编程实现下载
经常在论坛看见有人求Asp.Net中编程实现下载的代码,有些还希望能断点续传什么的.其实问题的关键在于权限.B/S和C/S不仅仅是外观上的区别而已. 下载,顾名思义是客户端要下,所以载.你硬塞給人家那 ...
- 【51Nod】1055 最长等差数列 动态规划
[题目]1055 最长等差数列 [题意]给定大小为n的互不不同正整数集合,求最长等差数列的长度.\(n \leq 10000\). [算法]动态规划 两个数之间的差是非常重要的信息,设\(f_{i,j ...
- [转]Restrict关键字
0 定义 C99中新增加的用于修饰指针的关键字,用于表示该指针所指向的内存,只有通过该指针访问得到(如下ptr指向的内存单元只能通过ptr访问得到).从而可以让编译器对代码进行优化,生成更有效率的汇编 ...
- Shell中各种判断语法
Shell判断 按照文件类型进行判断 -b 判断文件是否存在,并且是否为快设备文件(是块设备文件为真) -c 判断文件是否存在,并且是否为字符设备文件(是字符设备文件为真) -d 判断文件是否存在,并 ...
- 【Python】如何基于Python写一个TCP反向连接后门
首发安全客 如何基于Python写一个TCP反向连接后门 https://www.anquanke.com/post/id/92401 0x0 介绍 在Linux系统做未授权测试,我们须准备一个安全的 ...
- vim下如何去掉windows编辑的文件中的^M
可以去掉^M, 例如: 在终端下敲命令: dos2unix a.c 直接转换成unix格式,这样就可以去掉^M •$dos2unix filename •vim filename :%s/^M$//g ...