杭电-1003 Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 217222    Accepted Submission(s): 51237

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
  
  

Author

Ignatius.L

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#include <stdio.h>
int a[100010];
int main(){
	int T,N,i,cas=1;
	int start,end,temp,sum,max=0;
	scanf("%d",&T);
	while(T--){
		scanf("%d",&N);
		start=end=temp=1;
		sum=0;
		max=-1001;
		for(i=1;i<=N;i++){
			scanf("%d",&a[i]);
			sum+=a[i];
			if(sum>max)
			{
				max=sum;
				start=temp;
				end=i;
			}
			if(sum<0)
			{
				sum=0;
				temp=i+1;
			}
		}
		printf("Case %d:\n",cas++);
		printf("%d %d %d\n",max,start,end);
		if(T>0)
			printf("\n");
	}
	return 0;
}