hdu1003 Max Sum(dp)

Problem Description
Given a sequence a[1],a[2],a[3]……a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14.

Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000).

Output
For each test case, you should output two lines. The first line is “Case #:”, # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases.

Sample Input
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5

Sample Output
Case 1:
14 1 4

Case 2:
7 1 6

#include<cstdio>
#include<iostream>
#include<string.h>
using namespace std;
const int maxn=100000+5;
int dp[maxn];
int l[maxn];
int r[maxn];

int a[maxn];

int main(){
    int t;
    scanf("%d",&t);
    int cas=1;
    while(t--){
        memset(l,0,sizeof(l));
        memset(r,0,sizeof(r));
        printf("Case %d:\n",cas++);
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            dp[i]=a[i];
        } 
        l[1]=1,r[1]=1;
        for(int i=2;i<=n;i++){
            if(dp[i]<=dp[i-1]+a[i]){
                dp[i]=dp[i-1]+a[i];
                l[i]=l[i-1];
                r[i]=i;
            }
            else{
                l[i]=i,r[i]=i;
            }
            //printf("dp[%d]=%d, l=%d, r=%d\n",i,dp[i],l[i],r[i]);
        }
        int ans=dp[1];
        int i,j=1;
        for(i=2;i<=n;i++){
            if(ans<dp[i]){
                ans=dp[i];
                j=i;
            }

        }
        printf("%d %d %d\n",ans,l[j],r[j]);
        if(t)
            printf("\n");

    }
}