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：
用二分法（避免超时）找最小的数放在子序列里面，比如最开始的序列是 2 3 4，但是后面又发现了一个1，就把1代替之前的2，以免后面出现更长的子序列
1 #include<iostream>
2 #include<cstdio>
3 using namespace std;
4 #define maxn 100000+10
5 int al[maxn], cl[maxn];
6 int main()
7 {
8 int n,len,left,right,mid;
9 while (scanf("%d",&n)!= EOF){
10 for (int i = 0; i < n; i++)
11 scanf("%d", &al[i]);
12 len = 0, cl[0] = -1;
13 for (int i = 0; i < n; i++)
14 {
15 if (al[i]>cl[len])
16 cl[++len] = al[i]; //把大的数放进去
17 else
18 {
19 left = 1, right = len; //二分法找后面小的数放在这个子序列的前面，以免有更长的子序列
20
21 while (left <= right)
22 {
23 mid = (left + right) / 2;
24 if (al[i]>cl[mid])
25 left = mid + 1;
26 else
27 right = mid - 1;
28 }
29 cl[left] = al[i];
30 }
31 }
32 printf("%d\n", len);
33 }
34
35 return 0;
36 }