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
【题意】 给n个数，求最大连续元素的和；并输出起点和终点。
【分析】
设d[i]为以第i个元素为终点的最大连续元素和。则
状态转移方程：d[i] = d[i-1] > 0 ? d[i-1]+a[i] : a[i];
思路应该比较好理解，如果d[i-1]<0， 那么无论a[i]为何值，其和总不如单独的a[i]大；反之如果d[i-1]>0， 那么无论a[i]为何值，其和总大于单独的a[i]；
【代码】
1 #include<iostream>
2 #include<cstdio>
3 #include<cstdlib>
4 #include<cstring>
5 using namespace std;
6 const int maxn = 100010;
7 int n, a[maxn];
8 int d[maxn];
9 void dp()
10 {
11 memset(d, 0, sizeof(d));
12 d[0] = a[0];
13 for(int i = 1; i < n; i++)
14 {
15 if(d[i-1] > 0) d[i] = d[i-1]+a[i];
16 else d[i] = a[i];
17 }
18 //cout << endl;
19 int st = 0, en = 0;
20 int max_ = d[0];
21 for(int i = 1; i < n; i++)
22 {
23 if(d[i]>max_)
24 {
25 max_ = d[i];
26 en = i;
27 }
28 }
29 st = en;
30 for(int j = en-1; j>=0; j--)
31 {
32 if(d[j] < 0) break;
33 else st = j;
34 }
35 printf("%d %d %d\n", max_, st+1, en+1);
36 }
37
38 int main()
39 {
40 int T; scanf("%d", &T);
41 for(int kase = 0; kase < T; kase++)
42 {
43 scanf("%d", &n);
44 for(int i = 0; i < n; i++)
45 scanf("%d", &a[i]);
46 if(kase) printf("\n");
47 printf("Case %d:\n", kase+1);
48 dp();
49
50 }
51 return 0;
52 }