C. Ilya and Sticks
time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

In the evening, after the contest Ilya was bored, and he really felt like maximizing. He remembered that he had a set of n sticks and
an instrument. Each stick is characterized by its length li.

Ilya decided to make a rectangle from the sticks. And due to his whim, he decided to make rectangles in such a way that maximizes their total area. Each stick is used in making at most one rectangle, it is possible that some of sticks remain unused. Bending
sticks is not allowed.

Sticks with lengths a1, a2, a3 and a4 can
make a rectangle if the following properties are observed:

  • a1 ≤ a2 ≤ a3 ≤ a4
  • a1 = a2
  • a3 = a4

A rectangle can be made of sticks with lengths of, for example, 3 3 3 3 or 2 2 4 4.
A rectangle cannot be made of, for example, sticks5 5 5 7.

Ilya also has an instrument which can reduce the length of the sticks. The sticks are made of a special material, so the length of each stick can be reduced by at most one. For example, a stick with length 5 can
either stay at this length or be transformed into a stick of length 4.

You have to answer the question — what maximum total area of the rectangles can Ilya get with a file if makes rectangles from the available sticks?

Input

The first line of the input contains a positive integer n (1 ≤ n ≤ 105) — the
number of the available sticks.

The second line of the input contains n positive integers li (2 ≤ li ≤ 106) — the
lengths of the sticks.

Output

The first line of the output must contain a single non-negative integer — the maximum total area of the rectangles that Ilya can make from the available sticks.

Sample test(s)
input
4
2 4 4 2
output
8
input
4
2 2 3 5
output
0
input
4
100003 100004 100005 100006
output
10000800015

题目链接:点击打开链接

给出n个棒子的长度, 每一个棒子能够变为自己的len - 1, 问能够组成题目中要求的矩形的面子最大是多少.

将棒子排序, 由大到小遍历棒子, 因为每一个棒子能够变成自己的len - 1, 若相邻棒子差值为1或0, 则四边构成一个矩形, 累加得到答案.

AC代码:

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "queue"
#include "stack"
#include "cmath"
#include "utility"
#include "map"
#include "set"
#include "vector"
#include "list"
#include "string"
#include "cstdlib"
using namespace std;
typedef long long ll;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e6 + 5;
int n;
ll x, ans;
int main(int argc, char const *argv[])
{
cin >> n;
vector<int> v(n);
for(int i = 0; i < n; ++i)
cin >> v[i];
sort(v.begin(), v.end());
for(int i = n - 2; i >= 0; --i)
if(v[i + 1] - v[i] < 2) {
if(x) {
ans += x * v[i];
x = 0;
}
else x = v[i];
i--;
}
cout << ans << endl;
return 0;
}

Codeforces Round #297 (Div. 2) 525C Ilya and Sticks(脑洞)的更多相关文章

  1. Codeforces Round #297 (Div. 2)C. Ilya and Sticks 贪心

    Codeforces Round #297 (Div. 2)C. Ilya and Sticks Time Limit: 2 Sec  Memory Limit: 256 MBSubmit: xxx  ...

  2. 贪心 Codeforces Round #297 (Div. 2) C. Ilya and Sticks

    题目传送门 /* 题意:给n个棍子,组成的矩形面积和最大,每根棍子可以-1 贪心:排序后,相邻的进行比较,若可以读入x[p++],然后两两相乘相加就可以了 */ #include <cstdio ...

  3. Codeforces Round #297 (Div. 2)E. Anya and Cubes 折半搜索

    Codeforces Round #297 (Div. 2)E. Anya and Cubes Time Limit: 2 Sec  Memory Limit: 512 MBSubmit: xxx  ...

  4. Codeforces Round #297 (Div. 2)D. Arthur and Walls 暴力搜索

    Codeforces Round #297 (Div. 2)D. Arthur and Walls Time Limit: 2 Sec  Memory Limit: 512 MBSubmit: xxx ...

  5. Codeforces Round #297 (Div. 2)B. Pasha and String 前缀和

    Codeforces Round #297 (Div. 2)B. Pasha and String Time Limit: 2 Sec  Memory Limit: 256 MBSubmit: xxx ...

  6. Codeforces Round #297 (Div. 2)A. Vitaliy and Pie 水题

    Codeforces Round #297 (Div. 2)A. Vitaliy and Pie Time Limit: 2 Sec  Memory Limit: 256 MBSubmit: xxx  ...

  7. BFS Codeforces Round #297 (Div. 2) D. Arthur and Walls

    题目传送门 /* 题意:问最少替换'*'为'.',使得'.'连通的都是矩形 BFS:搜索想法很奇妙,先把'.'的入队,然后对于每个'.'八个方向寻找 在2*2的方格里,若只有一个是'*',那么它一定要 ...

  8. 字符串处理 Codeforces Round #297 (Div. 2) B. Pasha and String

    题目传送门 /* 题意:给出m个位置,每次把[p,len-p+1]内的字符子串反转,输出最后的结果 字符串处理:朴素的方法超时,想到结果要么是反转要么没有反转,所以记录 每个转换的次数,把每次要反转的 ...

  9. 模拟 Codeforces Round #297 (Div. 2) A. Vitaliy and Pie

    题目传送门 /* 模拟:这就是一道模拟水题,看到标签是贪心,还以为错了呢 题目倒是很长:) */ #include <cstdio> #include <algorithm> ...

随机推荐

  1. 最小生成树(Kruskal)(并查集)

    最小生成树 时间限制: 1 Sec  内存限制: 64 MB提交: 11  解决: 2[提交][状态][讨论版] 题目描述 某个宇宙帝国有N个星球,由于宇宙的空间是三维的,因此每个星球的位置可以用三维 ...

  2. Jenkins一个任务下载多个git库代码

    公司的项目是微服务架构,一个服务对应的一个git仓库,现在的需求时拉取所有仓库代码下来,指定父级的pom.xml,一次性构建打包 jenkins在默认情况下,一个任务只能配置一个git仓库地址 1.安 ...

  3. Java的ClassLoader机制

    http://blog.chenlb.com/2009/06/java-classloader-architecture.html http://blog.csdn.net/lovingprince/ ...

  4. luogu P1064 金明的预算方案

    题目描述 金明今天很开心,家里购置的新房就要领钥匙了,新房里有一间金明自己专用的很宽敞的房间.更让他高兴的是,妈妈昨天对他说:“你的房间需要购买哪些物品,怎么布置,你说了算,只要不超过N元钱就行”.今 ...

  5. 【分块】【树状数组】bzoj3787 Gty的文艺妹子序列

    题解懒得自己写了,Orz一发wangxz神犇的: http://bakser.gitcafe.com/2014/12/04/bzoj3787-Gty%E7%9A%84%E6%96%87%E8%89%B ...

  6. Exercise01_07

    public class Outcome{ public static void main(String[] args){ double x, y; x=4*(1.0-1.0/3+1.0/5-1.0/ ...

  7. 【R实践】时间序列分析之ARIMA模型预测___R篇

    时间序列分析之ARIMA模型预测__R篇 之前一直用SAS做ARIMA模型预测,今天尝试用了一下R,发现灵活度更高,结果输出也更直观.现在记录一下如何用R分析ARIMA模型. 1. 处理数据 1.1. ...

  8. 【OpenJudge9267】【递推】核电站

    核电站 总时间限制: 5000ms 单个测试点时间限制: 1000ms 内存限制: 131072kB [描述] 一个核电站有N个放核物质的坑,坑排列在一条直线上.如果连续M个坑中放入核物质,则会发生爆 ...

  9. Mybatis通过ID查询 && 通过name模糊查询

    接上篇:Mybatis环境搭建 在搭建环境时已经有了mapper和sqlMapConfig 1,数据库建表 prompt PL/SQL Developer import file prompt Cre ...

  10. Android studio百度地图demo出现230错误,key校验失败

    转自daoxiaomianzi原文 Android studio 百度地图demo出现230错误,key校验失败 使用AndroidStudio导入Baidu地图的as版的demo,引入后,发现没有k ...