HDU Max Sum



http://acm.hdu.edu.cn/showproblem.php?pid=1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 136938    Accepted Submission(s): 31734

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
  

 

去年暑假的时候做了这个题目，当时做的稀里糊涂的，应该只是套用模板写的，现在整理一下，对dp又有了更深层次的理解。

附上今年写的代码，这个要标记位置要注意一下，之前写的是在nyoj上，hdu的格式要求要高一点。

#include<stdio.h>
int a[100010];
int main(){
	int ncase,k=0,i,n,pos,end,thissum,maxsum,begin;
	scanf("%d",&ncase);
	while(ncase--){
		k++;
		scanf("%d",&n);
		for(i = 0;i < n;i++)
			scanf("%d",&a[i]);
		pos = begin = end = 0;
		maxsum = thissum = a[0];
		for(i = 1;i < n;i++){
			if(thissum + a[i] < a[i]){             //如果当前值比a[i]小的话则改为a[i]
				thissum = a[i];
				pos = i;                  //记录下改的位置
			}
			else{
				thissum = thissum + a[i];
			}
			if(thissum > maxsum){                //当前值比最大值大，则头尾都要改
				maxsum = thissum;
				begin = pos;
				end = i;
			}
		}
		printf("Case %d:\n%d %d %d\n",k,maxsum,begin+1,end+1);
		if(ncase)
			printf("\n");
	}
	return 0;
}