HDU 1003 (最大子序列和)

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

 

Sample Output

Case 1:14 1 4Case 2:
7 1 6
#include <stdio.h>
int main(){int t,j;scanf("%d",&t);    for(j=1;j<=t;j++){int n,i;int sum=0,max=-2000,start=1,temp=1,end;scanf("%d",&n);int a[n+1];for(i=1;i<=n;i++){   scanf("%d",&a[i]);     sum+=a[i]; if(max<=sum)   {   start=temp;     max=sum;     end=i;   }    if(sum<0){   sum=0;   temp=i+1;    }}printf("Case %d:\n%d %d %d\n",j,max,start,end);if(j!=t)printf("\n");}}