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
思路：最大子段和是线性dp的入门题目，重要的还是理解。求区间就是按照那个思路，相应的更新区间就可以了。具体看代码：
#include<bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
const int maxx=1e5+100;
int a[maxx];
int n;
int main()
{
int t;
scanf("%d",&t);
int k=0;
while(t--)
{
scanf("%d",&n);
for(int i=1;i<=n;i++) scanf("%d",&a[i]);
int ans=a[1],sum=-inf;
int L=1,R=1,l=1,r=1;
for(int i=2;i<=n;i++)
{
if(ans>=0) ans+=a[i],r++;
else
{
if(sum<ans) sum=ans,L=l,R=r;
ans=a[i],l=r=i;
}
if(sum<ans) sum=ans,L=l,R=r;
}
printf("Case %d:\n",++k);
printf("%d %d %d\n",sum,L,R);
if(t)puts("");
}
return 0;
}
努力加油a啊，(o)/~
