线性时间求解最大子序列和——HDU1003

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

#include using namespace std; int main(){ long cnt,n; long a[100000]; long temp_sum, ts, te; long max_sum, start, end; int i,j; j = 0; cin >> cnt; while(cnt--){ cin >> n; ++j; temp_sum = max_sum = -1024; for(i=0; i> a[i]; } for(i=0; i temp_sum){ temp_sum = a[i]; ts = te = i; if(temp_sum > max_sum){ max_sum = temp_sum; start = ts; end = te; } } }else{ temp_sum += a[i]; te = i; if(temp_sum > max_sum){ max_sum = temp_sum; start = ts; end = te; } } } cout << "Case " << j << ":" << endl; cout << max_sum << " " << start+1 << " " << end+1 << endl; if(cnt) cout << endl; } return 0; }