1003 Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 353912    Accepted Submission(s): 84524

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

 

AC代码：

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int main()
{
    int n, m, val, i, sum, j ,summax,left,rigth,tmp;
    vector<int> v;
    while (cin>>n)
    {
        for (i = 0; i < n; i++) {
            cin >> m;
            v.clear();
            for (j = 0; j < m; j++) {
                cin >> val;
                v.push_back(val);
            }
            sum = 0;
            summax = INT_MIN;
            left = rigth = 0;
            tmp = 0;
            for (j = 0; j < m; j++) {
                sum +=  v[j];
                if (sum > summax) {
                    summax = sum;
                    rigth = j;
                    left = tmp;
                }
                if (sum < 0) {
                    sum = 0;
                    tmp = j + 1;
                }
                
            }
            cout << "Case " << i+1<< ":" << endl;
            cout << summax << " " << left +1<< " " << rigth +1 << endl;
            if (i != n - 1) {
                cout << endl;
            }
        }
    }
    return 0;
}