Max Sum（经典DP）

Max Sum Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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

设一数组sum，sum[i]含义：所有以a[i]为结尾的序列的序列和构成一个集合，此集合的最大值就是sum[i]
如1,2,3.所有以a[2]=3为结尾的序列的序列和集合是{6,5,3},因而sum[2]=6.

sum的状态转移方程：sum[i] = max{sum[i-1]+a[i], a[i]}
ans必定是sum[0···(k-1)]之一。由于要记录起始位置和结束位置，引入s数组记录获得sum的序列的起始元素的位置，而由sum的定义，sum[i]的结束位置是i不用另外记录。

 1 //Memory: 1420 KB         Time: 0 MS
 2 //Language: C++         Result: Accepted
 3 #include <iostream>
 4 #include <string>
 5 #include <set>
 6 #include <map>
 7 #include <vector>
 8 #include <stack>
 9 #include <queue>
10 #include <cmath>
11 #include <cstdio>
12 #include <cstring>
13 #include <algorithm>
14 using namespace std;
15 #define LL long long
16 #define cti const int
17 #define ctll const long long
18 #define dg(i) cout << "*" << i << endl;
19 
20 int a[100005], sum[100005], s[100005];
21 
22 int main()
23 {
24     int T, N;
25     int ca = 0, ans;
26     scanf("%d", &T);
27     while(ca < T)
28     {
29         scanf("%d", &N);
30         for(int i = 0; i < N; i++)
31             scanf("%d", &a[i]);
32         ans = 0;
33         sum[0] = a[0];
34         s[0] = 0;
35         for(int i = 1; i < N; i++)
36         {
37             if(sum[i-1] >= 0)
38             {
39                 sum[i] = sum[i-1] + a[i];
40                 s[i] = s[i-1];
41             }
42             else
43             {
44                 sum[i] = a[i];
45                 s[i] = i;
46             }
47             if(sum[ans] < sum[i]) ans = i;
48         }
49         printf("Case %d:\n%d %d %d\n", ++ca, sum[ans], s[ans] + 1, ans + 1);
50         if(ca != T) puts("");
51     }
52     return 0;
53 }