Max Sum（lis）

A - Max Sum

Time Limit:1000MS    Memory Limit:32768KB    64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 1003

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

AC 代码：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int MAXN = 1e5+10;
int m[MAXN];
int t[MAXN];

int main()
{
    int f, k = 1, low = 0, high = 0;

    scanf("%d", &f);

    while (f--)
    {
        memset(m, 0, sizeof(m));
        memset(t, 0, sizeof(t));

        int n;

        scanf("%d", &n);

        for (int i=0; i<n; i++)
        {
            scanf("%d", &m[i]);
        }

        int ans, ii = -1;

        for(int i=0; i<n; i++)
        {
            if (m[i] >= 0)
            {
                ans = m[i];
                low = i;
                high = i;
                ii = i;
                break;
            }
        }

        if (-1 == ii)
        {
            ans = m[0];
            low = 0;
            high = 0;
            for (int i=1; i<n; i++)
            {
                if (ans < m[i])
                {
                    ans = m[i];
                    low = i;
                    high = i;
                }
            }
        }
        else
        {
            for (int i=ii; i<n; i++)
            {
                if (m[i] < 0)
                {
                    continue;
                }
                t[i] = m[i];

                if (ans < m[i])
                {
                    ans = m[i];
                    low = i;
                    high = i;
                }

                for (int j=i+1; j<n; j++)
                {

                    t[i] += m[j];

                    if (t[i] < 0)
                        break;

                    if (ans < t[i])
                    {
                        ans = t[i];
                        low = i;
                        high = j;
                    }
                }
            }
        }
        printf("Case %d:\n", k++);
        printf("%d %d %d\n", ans, low+1, high+1);

        if (f)
        {
            printf("\n");
        }

    }
    return 0;
}

这道题刷了好几天终于过了，毕竟是在瓶颈期间，据说是lis题，我没联想到，我的代码应该是dp吧，哈哈哈。