1829: Candies

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本文介绍了一个关于儿童如何根据特定规则公平分配编号糖果的问题，并提供了一段C++代码实现，该算法通过动态规划找到能获取的最大数量的糖果。

1829: Candies

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	816	179	Standard

All children like candies. One day a group of children are dividing M candies by the following rules.

1. The candies are numbered with 1,2,3…M.

2. One children can select a series of candies from small number to high number.

3. The candy chosen for the first time can be taken away.

4. From the second candy, it can be taken away iff its value is not more than the value of previous one.

Everyone wants to get candies as many as possible. Now assume you are the first child to choose candies, you are to write a program to calculate the largest number of candies you can take.

Input

There may be more than one test case in the input.

The first line of each case is an integer M(M<=1000), number of candies to be divided, followed by M lines indicating values of every candies. They are sorted by their numbers from 1 to M.

Output

For each case, print the answer.

Sample Input

5
5
2
4
1
3

Sample Output

This problem is used for contest: 148

#include<iostream>
//#include<cstring>
using namespace std;
int dp[1001];
int a[1001];
int main()
{
    int m,i,j;
    while(scanf("%d",&m)==1)
    {
        //memset(dp,0,sizeof(dp));
        for(i=1;i<=m;i++)
        {
            cin>>a[i];
            dp[i]=1;
        }
        //dp[0]=1;
        for(i=m;i>=1;i--)
        {
            for(j=i+1;j<=m;j++)
            {
                if(a[i]>=a[j]&&dp[i]<dp[j]+1)
                {
                    dp[i]=dp[j]+1;
                }
            }
        }
        int maxn=0;
        for(i=1;i<=m;i++)
        {
            if(dp[i]>maxn)
            {
                maxn=dp[i];
            }
        }
        cout<<maxn<<endl;
    }
    return 0;
}

//8 6 5 9 3 5 7 6 9