HDU - 1231 largest contiguous subsequence (basic dp)

Recently title suddenly found vj brush also has a few long-ago attempt problem is not resolved, so when the water problem to make up these questions ....

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Meaning of the questions:

Seeking a maximum contiguous subsequence sequence analysis, while its output range (the maximum value if there is the same section, the output i, j minimum interval)

Ideas:

Seeking a maximum contiguous subsequence: Recurrence Formula dp [i] = max (arr [i], dp [i-1] + arr [i]), represents a maximum contiguous subsequence end and i. Finally, the maximum output again traversed dp value.

About the interval range we write a structure records what you can, be updated when recursive.

code:

#include<cstdio>
#include<algorithm>
#include<iostream>
using namespace std;
#define maxn 10005
const int inf = 0x3f3f3f3f;
struct node{
    int l,r;
    int val;
}dp[maxn];
int arr[maxn];
int main()
{
    int n,mmax,flag;
    while(cin>>n&&n)
    {
        mmax=-inf , flag=0;
        for(int i=1;i<=n;i++) { 
            CIN >> ARR [I];
             IF (ARR [I]> = 0 ) = In Flag . 1 ; 
        } 
         // if all elements K are negative, and defines the maximum is 0, the output of the entire sequence of elements inclusive 
        IF (! In Flag) { 
            COUT << 0 << "  " << ARR [ . 1 ] << "  " << ARR [n-] << endl;
             Continue ; 
        } 
        // maximum continuous sequence DP 
        DP [ 0 ] .L = . 1 ;
         for ( int I = . 1 ; I <= n-; I ++){
            if(arr[i]>dp[i-1].val+arr[i]){
                dp[i].val = arr[i];
                dp[i].l = i;
                dp[i].r = i;
            }else{
                dp[i].val = dp[i-1].val + arr[i];
                dp[i].l = dp[i-1].l;
                dp[i].r = i;
            }
            mmax=max(dp[i].val,mmax);
        }
        int l,r;
        for(int i=1;i<=n;i++){
            if(dp[i].val==mmax){
                cout<<mmax<<" "<<arr[dp[i].l]<<" "<<arr[dp[i].r]<<endl;
                break;
            }
        }    
    }
    return 0;
}