hdu1087 最大递增子段和

Posted 2020-08-11 a_clown_cz

http://acm.split.hdu.edu.cn/showproblem.php?pid=1087

状态方程:sum[j]=max{sum[i]}+a[j]; 其中,0<=i<=j,a[i]<a[j] 

把当前最大和更新到数组中，注意顺序。

 

Input

Input contains multiple test cases. Each test case is described in a line as follow:
N value_1 value_2 …value_N 
It is guarantied that N is not more than 1000 and all value_i are in the range of 32-int.
A test case starting with 0 terminates the input and this test case is not to be processed.

 

 

Output

For each case, print the maximum according to rules, and one line one case.

 

 

Sample Input

3 1 3 2

4 1 2 3 4

4 3 3 2 1

0

Sample Output

4

10

3

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define INF 0xfffffff

int a[1005];
int dp[1005];

int main()
{
    int n;
    while(scanf("%d",&n)==1 && n)
    {
        for(int i=1; i<=n; i++)
            scanf("%d",&a[i]);
        memset(dp,0,sizeof(dp));
        int ans;
        for(int i=1; i<=n; i++)
        {
            ans=-INF;
            for(int j=0; j<i; j++)
                if(a[i]>a[j])
                    ans=max(ans,dp[j]);
            dp[i]=ans+a[i];
        }
        ans=-INF;
        for(int i=1; i<=n; i++)
            ans=max(ans,dp[i]);
        printf("%d\n",ans);
    }
    return 0;
}