poj_2479_MaximumSum.cpp

Problem Links:

poj2479,

Problem:

Maximum sum

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 20936		Accepted: 6347

Description

Given a set of n integers: A={a1, a2,..., an}, we define a function d(A) as below:

Your task is to calculate d(A).

Input

The input consists of T(<=30) test cases. The number of test cases (T) is given in the first line of the input.
Each test case contains two lines. The first line is an integer n(2<=n<=50000). The second line contains n integers: a1, a2, ..., an. (|ai| <= 10000).There is an empty line after each case.

Output

Print exactly one line for each test case. The line should contain the integer d(A).

Sample Input

1

10
1 -1 2 2 3 -3 4 -4 5 -5

Sample Output

13

Hint

In the sample, we choose {2,2,3,-3,4} and {5}, then we can get the answer.

Huge input,scanf is recommended.

Source

POJ Contest,Author:Mathematica@ZSU

Solution:

Source Code:

#include <stdio.h>
int main()
{
    freopen("input.in", "r", stdin);
    freopen("output.out", "w", stdout);
    int t, n, i, max, sum;
    int a[50000], b[50000], c[50000];
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d", &n);
        max = 0;
        sum = 0;
        if (n == 1)
        {
            scanf("%d", &a[0]);
            printf("%d\n", a[0]);
            continue;
        }
        if (n == 2)
        {
            scanf("%d", &a[0]);
            scanf("%d", &a[1]);
            printf("%d\n", a[0] + a[1]);
            continue;
        }
        //b[i] is the max sum from 0 to i.
        for (i = 0; i < n; i++)
        {
            scanf("%d", &a[i]);
            sum += a[i];
            if (sum < 0)
                sum = 0;
            else if (sum > max)
                max = sum;
            b[i] = max;
        }
        //c[i] is the max sum from i to the end.
        max = 0;
        sum = 0;
        for (i = n - 1; i >= 0; i--)
        {
            sum += a[i];
            if (sum < 0)
                sum = 0;
            else if (sum > max)
                max = sum;
            c[i] = max;
        }
        max = 0;
        for (i = 0; i < n - 1; i++)
            if (b[i] + c[i + 1] > max)
                max = b[i] + c[i + 1];
        printf("%d\n", max);
    }

    return (0);
}