hdoj1003-Max Sum(数组的最大和)

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

    题目的大致意思就是求有最大和的子序列，我的方法是边读取数据边处理，当和出现为负值是记录当前点的前一个为右端点，当和大于最大值是更新左右端点；    特别要注意的是数据可能全为负，因此max的值应初始化为-1005。

#include <iostream>#include <fstream>using namespace std;int main(){    //ifstream cin("data.in");    int t;    cin >> t;    int count = 1;    while(t --)    {        int n;        cin >> n;        int l = 0, l1 = 0, r = 0,  sum = 0, r1 = 0, max = -1005;               //数组有可能全为负,max初始值为-1005        for(int i = 0; i < n; i ++)        {            int temp;            cin >> temp;            sum += temp;            if(sum > max)            {      //出现最大值时更新右端点                l = l1;                r1 = i;                max = sum;            }            else if(sum < 0)            {     //当当前和为负数时，更新左端点                sum = 0;                l1 = i + 1;            }        }        if(count > 1)        {            cout << endl;        }        cout << "Case " << count++ << ":" << endl;        cout << max << " " << l+1 << " " << r1+1 << endl;    }    return 0;}