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

  1. 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 ...

  2. 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 ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. 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 ...

  7. 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 ...

  8. 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 ...

  9. 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

  1. phpStorm 8.0.3 设置

    phpstorm 8 license key Learn Programming===== LICENSE BEGIN =====63758-1204201000000Ryqh0NCC73lpRm!X ...

  2. HTML5 移动开发(移动设备检测及对HTML5的支持)

    1.如何选择要使用的特性以及所面向的浏览器 2.哪些浏览器支持HTML5 3.如何检测是否支持HTML5 4.如何开发贷容错性的Web应用程序 5.CSS3媒体查询如何增强检测脚本   使用HTML5 ...

  3. 第7月第17天 rxswift swift3.0

    1.rxswift just(...) .subscribe(onNext: { }) https://realm.io/cn/news/slug-max-alexander-functional-r ...

  4. python垃圾回收二

    由于循环引用的存在,我们在删除了a跟b之后,引用计数是1,这样,现有的垃圾回收机制是永远不可能把她们删除了.他们将永远存在于内存中. 我们当然不能对这种情况置之不理,于是,我们又添加了两种新的回收机制 ...

  5. 如何使用jpegtran 压缩JPG图片

    说到jpegtran相信很多人都比较陌生,网上相关的资料也很少. jpegtran可以让图片更加的简化,缩小图片的容量,从而增加网络的传输速度.说在多你也不信,下面就让事实证明. 首先下载  jpeg ...

  6. Linux进程托管与守护进程设置

    引言 在上一篇<Linux启动之旅>中,我们了解了Linux启动过程,在该过程的最后一步,init进程拉起/etc/init.d/rcN.d/目录下指定的守护进程(daemon).假若自定 ...

  7. linux,mac安装sentry

    linux,mac安装sentry 最近需要一个日志监视系统所以选择了sentry.以下是用mac安装,看需求量linux安装类似后面的文章会补充. 安装docker https://download ...

  8. MySQL安装与初步操作

    MySQL是一款出色的中小型关系数据库,做Java Web开发时,要做到数据持久化存储,选择一款数据库软件自然必不可少. 由于MySQL社区版开元免费,功能比较强大,在此以MySQL为例,演示MySQ ...

  9. ASP .Net Core系统部署到SUSE Linux Enterprise Server 12 SP3 64 具体方案

    .Net Core 部署到 SUSE Linux Enterprise Server 12 SP3 64 位中的步骤 1.安装工具 1.apache 2..Net Core(dotnet-sdk-2. ...

  10. R语言学习笔记:使用tcltk包显示进度条

    一般在跑耗时较长的程序时,我们不知道程序到底有没有正常跑着,或者在爬虫的时候不知道爬到什么时候断了.因此可以添加进度条来显示当前进度,观察进度是否有进展.当进度条卡住的时候,可以判断程序断线,从而可以 ...