D. Polycarp's Picture Gallery
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Polycarp loves not only to take pictures, but also to show his photos to friends. On his personal website he has recently installed a widget that can display n photos with the scroll option. At each moment of time the widget displays exactly one photograph with the option showing the previous/next one. From the first photo, you can switch to the second one or to the n-th one, from the second photo you can switch to the third one or to the first one, etc. Thus, navigation is performed in a cycle.

Polycarp's collection consists of m photo albums, the i-th album contains ai photos. Polycarp wants to choose n photos and put them on a new widget. To make watching the photos interesting to the visitors, he is going to post pictures so that no two photos from one album were neighboring (each photo will have exactly two neighbors, the first photo's neighbors are the second and the n-th one).

Help Polycarp compile a photo gallery. Select n photos from his collection and put them in such order that no two photos from one album went one after the other.

Input

The first line contains two integers n and m (3 ≤ n ≤ 1000, 1 ≤ m ≤ 40), where n is the number of photos on the widget, and m is the number of albums. The second line contains m integers a1, a2, ..., am (1 ≤ ai ≤ 1000), where ai is the number of photos in the i-th album.

Output

Print the single number -1 if there is no solution. Otherwise, print n numbers t1, t2, ..., tn, where ti represents the number of the album of the i-th picture in the widget. The albums are numbered from 1 in the order of their appearance in the input. If there are several solutions, print any of them.

Sample test(s)
Input
4 3
1 3 5
Output
3 1 3 2
Input
10 2
5 5
Output
2 1 2 1 2 1 2 1 2 1
Input
10 3
1 10 3
Output
-1
#include<cstdio>
#include<queue>
#include<algorithm>
using namespace std;
int a[],p[],co[];
typedef struct
{
int num,con;
}data;
data d[];
priority_queue<data>q;
bool operator<( data x, data y ){ return x.con<y.con;
}
int main()
{
int n,m,t,cnt,sum=,co,temp,right;
scanf("%d%d",&n,&m);
for(int i=; i<=m; i++)
{
scanf("%d",&a[i]);
if(a[i]>n/)
a[i]=n/;
sum+=a[i];
}
if(sum<n)
printf("-1");
else
{
for(int i=;i<=m;i++)
{
d[i].num=i;
d[i].con=a[i];
q.push(d[i]);
}
right=;
while(right<n)
{
data u=q.top();
q.pop();
for(int i=;i<=*u.con-&&(i+right<=n);i++)
if(i&)
p[i+right]=u.num;
else
{
data temp=q.top();
q.pop();
p[i+right]=temp.num;
temp.con--;
q.push(temp);
}
right=min(right+*u.con-,n);
}
for(int i=;i<=n;i++)
printf("%d ",p[i]); }
return ;
}

CodeForces 81D.Polycarp's Picture Gallery 乱搞的更多相关文章

  1. codeforces 653C C. Bear and Up-Down(乱搞题)

    题目链接: C. Bear and Up-Down time limit per test 2 seconds memory limit per test 256 megabytes input st ...

  2. Codeforces Gym 100203G Good elements 暴力乱搞

    原题链接:http://codeforces.com/gym/100203/attachments/download/1702/statements.pdf 题解 考虑暴力的复杂度是O(n^3),所以 ...

  3. Playrix Codescapes Cup (Codeforces Round #413, rated, Div. 1 + Div. 2)(A.暴力,B.优先队列,C.dp乱搞)

    A. Carrot Cakes time limit per test:1 second memory limit per test:256 megabytes input:standard inpu ...

  4. Codeforces 1182D Complete Mirror 树的重心乱搞 / 树的直径 / 拓扑排序

    题意:给你一颗树,问这颗树是否存在一个根,使得对于任意两点,如果它们到根的距离相同,那么它们的度必须相等. 思路1:树的重心乱搞 根据样例发现,树的重心可能是答案,所以我们可以先判断一下树的重心可不可 ...

  5. Codeforces 538G - Berserk Robot(乱搞)

    Codeforces 题目传送门 & 洛谷题目传送门 一道很神的乱搞题 %%% 首先注意到如果直接去做,横纵坐标有关联,不好搞.这里有一个非常套路的技巧--坐标轴旋转,我们不妨将整个坐标系旋转 ...

  6. Codeforces 306D - Polygon(随机化+乱搞)

    Codeforces 题目传送门 & 洛谷题目传送门 中考终于结束了--简单写道题恢复下状态罢. 首先这一类题目肯定没法用一般的方法解决,因此考虑用一些奇淫的乱搞做法解决这道题,不难发现,如果 ...

  7. Codeforces 732e [贪心][stl乱搞]

    /* 不要低头,不要放弃,不要气馁,不要慌张 题意: 给n个插座,m个电脑.每个插座都有一个电压,每个电脑都有需求电压. 每个插座可以接若干变压器,每个变压器可以使得电压变为x/2上取整. 有无限个变 ...

  8. Codeforces 1077E (二分乱搞或者dp)

    题意:给你一个数组,可以从中选区若干种元素,但每种元素选区的个数前一种必须是后一种的2倍,选区的任意2种元素不能相同,问可以选取最多的元素个数是多少? 思路1(乱搞):记录一下每种元素的个数,然后暴力 ...

  9. codeforces 665C C. Simple Strings(乱搞)

    题目链接: C. Simple Strings time limit per test 2 seconds memory limit per test 256 megabytes input stan ...

随机推荐

  1. python几种装饰器的用法

    用函数装饰函数 这种比较常见首先定义装饰器函数 def cache(func): data = {} @wraps(func) def wrapper(*args, **kwargs): key = ...

  2. python使用twisted搭建的一个socket服务

    服务端 # -*- coding: utf-8 -*- # @Time : 2018/9/19 21:41 # @Author : cxa # @File : tsTservTW.py # @Soft ...

  3. html5新增表单元素

    1.验证 <form> <input type="email"></input>    验证邮箱 <input type="ur ...

  4. sicily 1459. The Dragon of Loowater

            Time Limit: 1sec    Memory Limit:32MB  Description Once upon a time, in the Kingdom of Loowa ...

  5. HDU 6203 2017沈阳网络赛 LCA,DFS+树状数组

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=6203 题意:n+1 个点 n 条边的树(点标号 0 ~ n),有若干个点无法通行,导致 p 组 U V ...

  6. 19 Error handling and Go go语言错误处理

    Error handling and Go go语言错误处理 12 July 2011 Introduction If you have written any Go code you have pr ...

  7. ubuntu 16.04 网卡配置 虚拟机上网

    看所有网卡(包括未启动的) ifconfig -a 或者 ip link 查看当前网卡配置,打开配置文件/etc/network/interfaces 设置静态IP(dhcp 为动态获取,static ...

  8. python基础-实现进度条功能,for和yield实现

    实现进度条功能 方法一:简单FOR实现打印进度条功能 for i in range(10): print("#",end="",flush=True) time ...

  9. 简易博客[ html + css ] 练习

    1. 前言 通过使用 html + css 编写一个简易的博客作为入门练习 2. 代码及实现 2.1 目录结构 2.2 代码部分 <!DOCTYPE html> <html lang ...

  10. Linux下./configure && make && make install 编译安装和卸载

    正常的编译安装/卸载: 源码的安装一般由3个步骤组成:配置(configure).编译(make).安装(make install).   configure文件是一个可执行的脚本文件,它有很多选项, ...