练习3-A

题目：

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

题意：

求最大子段和；

思路：

输入数组的同时记录子段的和，并将当前子段的和与当前输入的数作比较，使子段和等于两者之间最大的数，当子段的和小于等于当前输入的数时，记录新的起点，并当新的最大子段和出现时，使用该新的起点。最初对这类题理解的不是很深刻，看着题中没有“如果某子序列全是负数.....”这个条件，就以为这个题和例题是不一样的，做完之后，仔细一想，不管这个条件有没有，都可以让子段和与0进行比较！

感想：

做题时的思路是正确的，但在输出起点和重点，尤其是起点的问题上浪费了太多太多太多的时间，虽然说这道题是最基础的。！！

一下午就搞出来了这一道题，饿死了，吃饭去！

代码：

# include 
# include 
# include 
using namespace std;

int max(int a, int b)
{
    return a>b? a:b;
}

int main()
{
    int T ,ccc = 0;
    cin >> T;

    for(int j = 1; j <= T; j++)
    {
        if (ccc)
            cout << endl;
        ccc = 1;
        int N;
        cin >> N;
        int a[100090], b[100090], be = 0, k = 0, maxx = 0;
        memset(a,0,sizeof(a));
        memset(b,0,sizeof(b));
        cin >> a[0];
        b[0] = a[0];
        for (int i = 1; i < N; i++)
        {
            cin >> a[i];
            b[i] = max(b[i-1]+a[i],a[i]);

            if (b[i] <= a[i] )
            {
                k = i;
            }

            if (b[i] > b[maxx])
            {
                maxx = i; //
                be = k;

            }
        }

        cout << "Case " <