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
分析：
DP题。状态转移方程：dp[i]=d[i-1]+a[i]>a[i]?d[i-1]+a[i]:a[i];
错误代码如下：（关键错误和正确代码的对比）
1 #include<iostream>
2 #include<cstdio>
3 using namespace std;
4 const int m=100005;
5 int a[m];
6
7 int main()
8 {
9 int t,k=0;
10 scanf("%d",&t);
11 while(t--)
12 {
13 int n,i,maxn,before,begin=1,end=1;
14 scanf("%d",&n);
15 for(i=1;i<=n;i++)
16 {
17
18 scanf("%d",&a[i]);
19 if(i==1)
20 {
21 maxn=a[i]; before=a[i];
22 }
23 else
24 {
25 if(before+a[i]<a[i])
26 {
27 before=a[i];
28 begin=i;
29 }
30 else
31 {
32 before+=a[i];
33 }
34
35 }
36 if(before>maxn)
37 {
38 maxn=before;
39 //x=begin;
40 end=i;
41 }
42
43 }
44 k++;
45 printf("Case %d:\n",k);
46 printf("%d %d %d\n",maxn,begin,end);
47 if(t)
48 printf("\n");
49
50 }
51
52 return 0;
53 }
正确代码如下：（用一个特殊点的案例6 :2 3 -6 4 3 -8找出错误！加入一个关键变量j记录当前a[i]位置）
1 #include<iostream>
2 #include<cstdio>
3 using namespace std;
4 const int m=100005;
5 int a[m];
6
7 int main()
8 {
9 int t,k=0;
10 scanf("%d",&t);
11 while(t--)
12 {
13 int n,i,maxn,before,begin=1,end=1,j=1; //必须用一个j记录当前数a[i]的位置
14 scanf("%d",&n);
15 for(i=1;i<=n;i++)
16 {
17
18 scanf("%d",&a[i]);
19 if(i==1)
20 {
21 maxn=a[i]; before=a[i];
22
23 }
24 else
25 {
26 if(before+a[i]<a[i])
27 {
28 before=a[i];
29 j=i; //记录当前位置，此时begin不改变
30 }
31 else
32 {
33 before+=a[i];
34 }
35
36 }
37 if(before>maxn)
38 {
39 maxn=before;
40 begin=j; //只有当before>maxn时再改变
41 end=i;
42 }
43
44 }
45 k++;
46 printf("Case %d:\n",k);
47 printf("%d %d %d\n",maxn,begin,end);
48 if(t)
49 printf("\n");
50
51 }
52
53 return 0;
54 }