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: ",++Case); 40 printf("%lld %d %d ",sum,num[q],q); 41 if(t) printf(" "); 42 } 43 return 0; 44 }