Codeforces Round #260 (Div. 1) A. Boredom
2015-08-14 23:38
447 查看
http://codeforces.com/problemset/problem/455/A
A. Boredom
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Alex doesn’t like boredom. That’s why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it.
Given a sequence a consisting of n integers. The player can make several steps. In a single step he can choose an element of the sequence (let’s denote it ak) and delete it, at that all elements equal to ak + 1 and ak - 1 also must be deleted from the sequence. That step brings ak points to the player.
Alex is a perfectionist, so he decided to get as many points as possible. Help him.
Input
The first line contains integer n (1 ≤ n ≤ 105) that shows how many numbers are in Alex’s sequence.
The second line contains n integers a1, a2, …, an (1 ≤ ai ≤ 105).
Output
Print a single integer — the maximum number of points that Alex can earn.
Sample test(s)
Input
2
1 2
Output
2
Input
3
1 2 3
Output
4
Input
9
1 2 1 3 2 2 2 2 3
Output
10
Note
Consider the third test example. At first step we need to choose any element equal to 2. After that step our sequence looks like this [2, 2, 2, 2]. Then we do 4 steps, on each step we choose any element equals to 2. In total we earn 10 points.
分析:
题意:
给定一个序列,每次从序列中选出一个数ak,获得ak的得分,
同时删除序列中所有的ak−1,ak+1,
求最大得分的值。
思路:
存下每个数的个数放在c中,消除一个数i,会获得c[i]*i的值(因为可以消除c[i]次),
如果从0的位置开始向右消去,那么,消除数i时,i-1可能选择了消除,也可能没有,
如果消除了i-1,那么i值就已经不存在,dp[i] = dp[i-1],
如果没有被消除,那么dp[i] = dp[i-2]+ c[i]*i。
A. Boredom
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Alex doesn’t like boredom. That’s why whenever he gets bored, he comes up with games. One long winter evening he came up with a game and decided to play it.
Given a sequence a consisting of n integers. The player can make several steps. In a single step he can choose an element of the sequence (let’s denote it ak) and delete it, at that all elements equal to ak + 1 and ak - 1 also must be deleted from the sequence. That step brings ak points to the player.
Alex is a perfectionist, so he decided to get as many points as possible. Help him.
Input
The first line contains integer n (1 ≤ n ≤ 105) that shows how many numbers are in Alex’s sequence.
The second line contains n integers a1, a2, …, an (1 ≤ ai ≤ 105).
Output
Print a single integer — the maximum number of points that Alex can earn.
Sample test(s)
Input
2
1 2
Output
2
Input
3
1 2 3
Output
4
Input
9
1 2 1 3 2 2 2 2 3
Output
10
Note
Consider the third test example. At first step we need to choose any element equal to 2. After that step our sequence looks like this [2, 2, 2, 2]. Then we do 4 steps, on each step we choose any element equals to 2. In total we earn 10 points.
分析:
题意:
给定一个序列,每次从序列中选出一个数ak,获得ak的得分,
同时删除序列中所有的ak−1,ak+1,
求最大得分的值。
思路:
存下每个数的个数放在c中,消除一个数i,会获得c[i]*i的值(因为可以消除c[i]次),
如果从0的位置开始向右消去,那么,消除数i时,i-1可能选择了消除,也可能没有,
如果消除了i-1,那么i值就已经不存在,dp[i] = dp[i-1],
如果没有被消除,那么dp[i] = dp[i-2]+ c[i]*i。
#include <iostream> #include <sstream> #include <iomanip> #include <vector> #include <deque> #include <list> #include <set> #include <map> #include <stack> #include <queue> #include <bitset> #include <string> #include <numeric> #include <algorithm> #include <functional> #include <iterator> #include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #include <cctype> #include <complex> #include <ctime> #define INF 0x3f3f3f3f #define eps 1e-6 #define p(x) printf("%d\n", x) #define k(x) printf("Case %d: ", ++x) #define mes(x, d) memset(x, d, sizeof(x)) #define s(x) scanf("%d", &x) typedef long long LL; const double pi = acos(-1.0); const long long mod = 1e9 + 7; using namespace std; LL dp[100005]; LL c[100005]; int main() { memset(dp,0,sizeof(dp)); memset(c,0,sizeof(c)); int N; scanf("%d",&N); LL t; LL MAX = -1; for(int i = 1;i <= N;i++) { scanf("%I64d",&t); MAX = max(MAX,t); c[t]++; } dp[0] = 0; dp[1] = a[1]; for(int i = 2;i <= MAX;i++) dp[i] = max(dp[i - 1],dp[i - 2] + i * c[i]); printf("%I64d\n",dp[MAX]); return 0; }
相关文章推荐
- 如何解决Mac里面解压后文件名乱码问题
- 上班一个月体会
- 【思维】uva11261Bishops
- JavaWeb开发常见乱码处理之设置编码方式
- C++模板(一)
- Eclipse插件安装
- Spring容器中Bean的实例化
- ListView,GridView和适配器Adapter不得不说的秘密
- 第0007道练习题_Python统计代码行数注释行数空白行数
- Linux C open打开文件,然后清空文件内容
- Mybatis的OneToMany,ManyToOne和解决N+1查询问题
- 在eclipse打开的android虚拟手机,打开File Explorer,下面是空的没有data、mnt、system三个文件
- 构建Tomcat 6
- js 函数
- 时间管理
- 黑马程序员—-C语言入门十重奏之八相念
- JavaEE细节问题02——加载资源文件的三种方式
- CS224d Problem set 3作业
- AnimalWindow使用,实现界面动态消失
- Web性能压力测试工具之WebBench详解