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. Asp.Net使用百度编辑器(ueditor)

    1.  1.4.3以上版本将不再承诺支持ie6/ie7. 2.如果是aspx 需要加上  ValidateRequest="false" 3.Web.config <syst ...

  2. angular 如何使用第三方组件ng-bootstrap

    1.在你的项目中以下指令    npm install --save @ng-bootstrap/ng-bootstrap 安装完成会显示 + @ng-bootstrap/ng-bootstrap@1 ...

  3. local class incompatible: stream classdesc serialVersionUID = -2897844985684768944, local class serialVersionUID = 7350468743759137184

    local class incompatible: stream classdesc serialVersionUID = 1, local class serialVersionUID = 2427 ...

  4. Android Studio配置opencv

    安装过程参考:http://www.cnblogs.com/tail/p/4618476.html demo参考:http://blog.csdn.net/gao_chun/article/detai ...

  5. 洛谷P2017晕牛

    传送门啦 这个题没有想象中复杂. 我们先有向边建立,因为我们无法改变有向边的方向. 建立完之后跑一边拓扑排序,我们按拓扑序小的指向大的就好了. 解释: 我们知道如果一个点在另一个点顺序的后面的话,如果 ...

  6. AdvStringGrid 固定行、列

  7. [扩展推荐] Laravel 中利用 GeoIP 获取用户地理位置信息

    我最近需要一个用户地域检测来设置用户的默认区域和货币.由 Daniel Stainback 创建的 torann/geoip  很好地满足为Laravel 5 项目提供 GeoIP 服务的要求. 这个 ...

  8. SQL中的left outer join,inner join,right outer join用法详解

    这两天,在研究SQL语法中的inner join多表查询语法的用法,通过学习,发现一个SQL命令,竟然涉及到很多线性代数方面的知识,现将这些知识系统地记录如下: 使用关系代数合并数据1 关系代数合并数 ...

  9. sicily 1198. Substring (递归全排列+排序)

    DescriptionDr lee cuts a string S into N pieces,s[1],…,s[N]. Now, Dr lee gives you these N sub-strin ...

  10. RabbitMQ系列之RabbitMQ单机安装

    安装epel源 rpm -ivh http://download.fedoraproject.org/pub/epel/6/x86_64/epel-release-6-8.noarch.rpm #ht ...