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Shi realized that he was almost out of money, even renting Shitalian lands. Shi was walking on a street, while thinking of a way to recover his fortune. In his way, he passed by a jewelry store. The owner of the store was a Shitalian man suspected of committing minor crimes, as cutting bushes and stealing old bread. Shi hated people who cut bushes, so he decided to rob the store to take revenge and make some money.
The store has n jewels, put in a row, one after another. Each jewel i can be sold in the black market for a value vi. Shi wants to steal as much as possible, but if he steals everything, the owner will notice, so if Shi steals the i-th jewel, he can‘t steal the i - 1-th, i - 2-th, i + 1-th and i + 2-th jewels.
Using the best strategy possible, calculate the sum of the jewels values that Shi can obtain.
The input begins with an integer n (1 ≤ n ≤ 106), indicating the number of jewels in the store. The next line containsn integers. The i-th integer vi (1 ≤ vi ≤ 103) is the value of the i-th jewel.
Output the maximum value that Shi can get.
4
1 2 3 4
5
6
1 2 4 0 3 0
5
7
2 10 12 24 29 69 0
81
10
15 1 6 3 7 100 9 15 80 95
210
题意：n个数字，每次最少隔两个取一个，求取得数的最大和
分析：dp，想法一、s[i]表示取第i个时最大和为多少，那就取不了i-1、i-2，可以取i-3、i-4、i-5，.....，当取i-6时，可取i-3，显然取了更划算，同理，i-7、i-8在算s[i-4]、s[i-5]时考虑过了，s[i]=a[i]+max{s[i-3],s[i-4],s[i-5]}
#include<stdio.h>
#include<algorithm>
using namespace std;
int n,s[1000025],tem,ans=0;
int main(){
scanf("%d",&n);
for(int i=5;i<n+5;i++)
{
scanf("%d",&tem);
s[i]=max(s[i-3],max(s[i-4],s[i-5]))+tem;
ans=max(ans,s[i]);
}
printf("%d",ans);
return 0;
}
想法二、s[i]表示前i个的最大和为多少，s[i]=max{s[i-1],s[i-3]+第i个}
#include<stdio.h>
#include<algorithm>
using namespace std;
int n,s[1000005],tem;
int main(){
scanf("%d",&n);
for(int i=3;i<n+3;i++)
scanf("%d",&tem),s[i]=max(s[i-1],tem+s[i-3]);
printf("%d",s[n+2]);
return 0;
}
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原文地址：http://www.cnblogs.com/flipped/p/5183142.html
