Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 207989 Accepted Submission(s): 48681

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

本题用DP求最大子序列，用l和r标记下标
用一个b记录子序列大小，若为负，则对整个序列起减小作用，不取，

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
using namespace std;

int main()
{
    int o,n,a[100005],b,mx,l,r,l1,r1;
    while(~scanf("%d",&o))
    { 
        int t=0;
    int q=o;
    while(o--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        b=0;
        mx=-1001;
        l=r=l1=r1=1;

        for(int i=1;i<=n;i++)
        {
            if(b<0)
            {
                l=i;
                r=i;
                b=a[i];
            }
            else
            {
                r=i;
                b+=a[i];
            }                    
            if(mx<b) 
            {
                l1=l;
                r1=r;
                mx=b;
            }
        }
        printf("Case %d:\n",++t);
        printf("%d %d %d\n",mx,l1,r1);
        if(t!=q)
            printf("\n");
    }
    }
    return 0;
}