最大连续子序列-杭电1003题

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题目：

Max Sum

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 51385 Accepted Submission(s): 11459

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



AC代码：

#include<stdio.h>
int a[100002];
int main()
{
	freopen("IO.txt","r",stdin);
	int NowMax,Nowstart,Nowend; //当前的最大值，开始位置，终止位置
	int Max,start,end;	    //最大值，开始位置，终止位置
	int i,j,T,n;
	while(scanf("%d",&T)!=EOF)
	{
		for(i=0;i<T;i++)
		{
			scanf("%d",&n);
			for(j=0;j<n;j++)
			{
				scanf("%d",&a[j]);
			}
//			Max=-99999999;
			NowMax=Max=a[0];//让当前最大值和最大值全赋为数组的第一个值
			Nowstart=Nowend=start=end=0;
			for(j=1;j<n;j++)
			{
				if(NowMax>=0)//如果当前最大值大于等于0，加上当前的数组值，这里要注意不能忘记等号，如果没有等号开始位置就会不对
				{
					NowMax+=a[j];
				}
				else        //如果当前的序列的值小于0，那么就抛弃当前的序列，重新起一个序列
				{
					NowMax=a[j];
					Nowstart=j;
				}
				if(Max<NowMax)//判断最大值于当前序列最大值大小
				{
					Max=NowMax;
					start=Nowstart;//将开始位置赋给start
					end=j;
				}
			}
			printf("Case %d:/n",i+1);
			printf("%d %d %d/n",Max,start+1,end+1);
			if(i<T-1)
				printf("/n");
		}
	}
	return 0;
}