【杭电1003】Max Sum

Max Sum

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

#include<cstdio>
#include<algorithm>
using namespace std;
int main()
{
int t,n,a,k=1;
scanf("%d",&t);
for(int j=1;j<=t;j++)
{
	scanf("%d",&n);
	int f=1,start,to;
	long sum=0,max=-10001;//存在全为负数的情况，所以将最大值赋为负数 
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&a);
		sum+=a;
		if(sum>max)//更新最大值 
		{
			max=sum;
			start=f;//记录初位置 
			to=i;//记录末位置 
		}
		if(sum<0)//以一个正数开始 
		{
			sum=0;
			f=i+1;
		}
		
	}
	printf("Case %d:\n",k++);
	printf("%d %d %d\n",max,start,to);
	if(j<t)//注意最后没有空行 
	printf("\n");
	
}
	return 0;
}