Max Sum

点击打开链接

Problem 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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

 

Sample Output

Case 1:14 1 4Case 2:7 1 6

 

Author

Ignatius.L

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#define INF 0x3f3f3f3f
using namespace std;
int a[100005];
int main()
{int t,Case=1;
scanf("%d",&t);
while(t--)
{
int n,endd,sta;
scanf("%d",&n);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
int tst=1,ans=-INF,sum=0;
for(int i=1;i<=n;i++)
{
sum+=a[i];
if(sum>ans)
{
ans=sum;
sta=tst;
endd=i;
}
if(sum<0)
{
sum=0;
tst=i+1;
}
}
printf ("Case %d:\n",Case++);
printf("%d %d %d\n",ans,sta,endd);
if(t!=0)
printf("\n");
}
return 0;
 }