HDU1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 198961    Accepted Submission(s): 46506

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

 

最大字段和 状态转移方程:if(dp[i-1]>=0)　　dp[i]=dp[i-1]+a[i] 　　else dp[i]=a[i];

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
const int MAXN=100005;
int dp[MAXN];
int a[MAXN];
int n;
int main()
{
    int T;
    scanf("%d",&T);
    for(int cas=1;cas<=T;cas++)
    {
        scanf("%d",&n);    
        for(int i=0;i<n;i++)
            scanf("%d",&a[i]);
        int maxn=dp[0]=a[0];
        int l=0,r=0,ll=0,rr=0;
        for(int i=1;i<n;i++)
        {
            if(dp[i-1]>=0)
            {
                dp[i]=dp[i-1]+a[i];
                r++;
            }
            else
            {
                l=r=i;
                dp[i]=a[i];
            }
            if(dp[i]>maxn)
            {
                maxn=dp[i];
                ll=l;
                rr=r;
            }
        }
        printf("Case %d:
",cas);
        printf("%d %d %d
",maxn,ll+1,rr+1);
        if(cas!=T)
            printf("
");
    }
    
    return 0;
}