Max Sum

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

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
题意：已知若干组数，对于每一组数，求和最大的子序列，输出最大和，并输出子序列的起末位置；
代码：
#include<stdio.h>
int main()
{
	int a[100010],i,j,sum,t,p,q,s,k=0,n,maxsum;
	scanf("%d",&n);
	while(n--)
{
	scanf("%d",&t);
		for(i=1;i<=t;i++)
		scanf("%d",&a[i]);
		a[0]=1000; //必须定义，因为数组是从下标0开始的； 
		maxsum=a[1];
		p=1;q=1;sum=0;s=1;
		for(i=1;i<=t;i++)
	   {
		  sum+=a[i];
		  if(maxsum<sum)
		  {
		  	maxsum=sum;
		  	s=p;
		  	q=i;
		  }
		  if(sum<0)  
		  {
		  	sum=0;
		  	p=i+1;
		  }
	   }
	   printf("Case %d:\n",++k);
	   printf("%d %d %d\n",maxsum,s,q);
	   if(n>=1)
	   printf("\n");
	}
	return 0;
}