A - 最大子段和
Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u
Submit Status

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
 
解题思路：

题目意思：最大子段和是要找出由数组成的一维数组中和最大的连续子序列。比如{5,-3,4,2}的最大子序列就是 {5,-3,4,2}，它的和是8,达到最大；而 {5,-6,4,2}的最大子序列是{4,2}，它的和是6。看的出来了，找最大子序列的方法很简单，只要前i项的和还没有小于0那么子序列就一直向后扩展，否则丢弃之前的子序列开始新的子序列，同时我们要记下各个子序列的和，最后找到和最大的子序列

 
程序代码：
 1 #include <cstdio>
 2 using namespace std;
 3 const int N=100100;
 4 int a[N],n,b[N],num[N];
 5 long long sum;
 6 int q;
 7 int main()
 8 {
 9     int t,Case=0;
10     scanf("%d",&t);
11     while(t--)
12     {
13         scanf("%d",&n);
14         for(int i=1;i<=n;i++)
15                 scanf("%d",&a[i]);
16         b[1]=a[1];num[1]=1;
17         for(int i=2;i<=n;i++)
18         {
19             if(b[i-1]>=0)
20                {
21                   b[i]=b[i-1]+a[i];
22                   num[i]=num[i-1];
23                }
24              else
25              {
26                  b[i]=a[i];
27                  num[i]=i;
28              }
29         }
30         sum=b[1];q=1;
31         for(int i=2;i<=n;i++)
32         {
33             if(b[i]>sum)
34             {
35                 sum=b[i];
36                 q=i;
37             }
38         }
39         printf("Case %d:\n",++Case);
40         printf("%lld %d %d\n",sum,num[q],q);
41         if(t) printf("\n");
42     }
43     return 0;
44 }
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