线性时间求解最大子序列和——HDU1003

Problem Description

Givena sequence a[1],a[2],a[3]......a[n], your job is to calculate the maxsum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum inthis sequence is 6 + (-1) + 5 + 4 = 14.

 

 

Input

Thefirst line of the input contains an integer T(1<=T<=20) whichmeans the number of test cases. Then T lines follow, each line startswith a number N(1<=N<=100000), then N integers followed(all theintegers are between -1000 and 1000).

 

 

Output

Foreach test case, you should output two lines. The first line is "Case#:", # means the number of the test case. The second line containsthree integers, the Max Sum in the sequence, the start position of thesub-sequence, the end position of the sub-sequence. If there are morethan one result, output the first one. Output a blank line between twocases.

 

 

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 <iostream>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<n; ++i){cin >> a[i];}for(i=0; i<n; ++i){if(temp_sum < 0){if(a[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;}