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
思路分析：
在输入的同时就一边计算。
把当前输入的值都计算出来now+bef，再与刚刚输入进去的值比较now。如果now大，就把bef更新。并且记录此时的计算位置。
每次bef与max比较，直到找出最大值。
/(ㄒoㄒ)/~~终究摆脱不了百度，看别人博客。。。。
源代码：
1 #include<iostream> 2 using namespace std; 3 int main() 4 { 5 int T, icase = 0, max, n, bef, now, s, e, k; 6 cin >> T; 7 while (T--) 8 { 9 10 icase++; 11 cin >> n; 12 for (int i = 1; i <=n; i++) 13 { 14 cin >> now; 15 if (i == 1) 16 { 17 max=bef =now ; 18 k = s = e = 1; //初始化ing…… 19 } 20 else 21 { 22 if (now > bef + now)//当前值大于之前的和 23 { 24 bef = now; //把当前的值作为最新的值 25 k = i; //记录更新的开始位置 26 } 27 else 28 bef+=now; //计算总和 29 } 30 if (bef > max) //当前值大于了之前计算的最大值，max一直等于now 31 max = bef, s = k, e = i; 32 } 33 34 cout << "Case " << icase << ": "<< max << " " << s << " " << e << endl; 35 if (T) 36 cout << endl; 37 } 38 return 0;
39 }
心得：
加油加油加油↖(^ω^)↗干吧得！