hdu1003-Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 279144    Accepted Submission(s): 66274

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

 

这道题与“最大连续子序列和”类似，所以还是用类似的dp思想。

只是最大连续子序列和多了一个注：【注】：为方便起见，如果所有整数均为负数，则最大子序列和为0。

在讨论里找的多几组测试数据

4 0 0 2 0 —— 2 1 3
6 2 7 -9 5 4 3 —— 12 1 6
4 0 0 -1 0 —— 0 1 1
7 -1 -2 -3 -2 -5 -1 -2 —— -1 1 1
6 -1 -2 -3 1 2 3 —— 6 4 6
----------------------------------------------------------------------------------
增加一组测试数据：
5 -3 -2 -1 -2 -3 —— -1 3 3

 

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

#define INF 0x3f3f3f3f
#define MAXN 100010

int main()
{
    int T;
    cin >>T;
    int i, j;
    int a;
    for(i = 1;i <= T;++i)
    {
//      可以推断出当 t > 0 时可由 t += a 计算出最大值（就像最大连续子序列和）
//      当 t < 0 时，按照dp思想，dp[i] = max(dp[i-1] + a[i], a[i])
//      所以分两种情况，t < a 时，t = a;
//            
//      t > a 时，t += a（这并不影响最大值ans，例如-1，-2，ans已经存了-1）
//      为什么要 +=a？是为了保证子序列的连续，直到出现 t < a
        int n;
        cin >>n>>a;
        int t = a, l = 1, r = 1;
        int ans = a, num_start = 1, num_end = 1;
        for(j = 2;j <= n;++j)
        {
            cin >>a;  
            if(t < 0 && t < a)
            {
                t = a;
                l = r = j;
            }
            else
            {
                t += a;
                r++;
            }
            if(ans < t)
            {
                ans = t;
                num_start = l;
                num_end = r;
            }
        }
        cout <<"Case "<<i<<":"<<endl
            <<ans<<" "<<num_start<<" "<<num_end<<endl;
        if(i != T)
            cout <<endl;
    }
    return 0;
}