HDU 1003 动态规划 求最大子串和

首先是题目：大意是求最大子串和

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 126443    Accepted Submission(s): 29293

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

#include<stdio.h>
void main()
{
	int cases,i,m,n,sum,max,start,end,laststart,temp;
	while(scanf("%d",&cases)!=EOF)
	{
		for(m=1;m<=cases;m++)
		{
			scanf("%d",&n);
			sum=0;
			max=-1001;//*****
			start=end=laststart=0;
			for(i=0;i<n;i++)
			{	
				scanf("%d",&temp);
				sum+=temp;
                if(sum>max)
				{
                    start=laststart;
                    end=i;
                    max=sum;
                }
                if(sum<0)
				{
                    laststart=i+1;
                    sum=0;
                }
			}
			printf("Case %d:\n",m);
			printf("%d %d %d\n",max,start+1,end+1);
			if(m!=cases)
				printf("\n");
		}
	}
}

个人小结 1、设置一个laststart用来记录上一个开始的位置 2、sum设为0，max设为-1001，题目有数据范围-1000到1000