HDU1003——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 N,n;int max;int i,j,x;int sum,start,end;int a[100005];scanf("%d",&N);for(i=1;i<=N;i++){scanf("%d %d",&n,&a[1]);max=sum=a[1];start=end=x=1;for(j=2;j<=n;j++){scanf("%d",&a[j]);if(max+a[j]<a[j]){max=a[j];x=j;}else{max+=a[j];}if(sum<max){sum=max;start=x;end=j;}}printf("Case %d:\n",i);printf("%d %d %d\n",sum,start,end);if(i!=N)  {printf("\n");}}return 0;}