hdu 1003 Max Sum - dp
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（注意第一个5是表示一共5个数）
7 0 6 -1 1 -6 7 -5
Sample Output
Case 1: 14 1 4
Case 2: 7 1 6
思路：
注意考虑全为负的情况。
源代码：
1 #include <iostream>
2 #include<stdio.h>
3
4 using namespace std;
5
6 int main()
7 {
8 int cas,s,e,s1;
9 int j=1;
10 scanf("%d",&cas);
11 while(cas--)
12 {int sum=0,max=-100000;
13 int n,num;
14 scanf("%d",&n);
15 s1=1;
16 for(int i=1;i<=n;i++)
17 {
18
19 scanf("%d",&num);
20 sum+=num;
21 if(sum>max)
22 {max=sum;
23 s=s1;
24 e=i;}
25 if(sum<0)
26 {
27 sum=0;
28 s1=i+1;
29 }
30
31 }if(j-1)printf("\n");
32 printf("Case %d:\n",j++);
33 printf("%d %d %d\n",max,s,e);
34 }
35 return 0;
36 }
