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): 113459 Accepted Submission(s): 26237
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
#include<stdio.h>
# define M 100000
int main()
{
int i,n;
int a[M+5];
scanf("%d",&n);
for(i=1;i<=n;i++)
{
int j,k,q=1,m,max=-1001,s=0,e=0,sum=0;
scanf("%d",&m);
for(j=0;j<m;j++)
{
scanf("%d",&a[j]);
sum+=a[j];
if(max<sum)
{
max=sum;
s=q;
e=j+1;
}
if(sum<0)
{
sum=0;
q=j+2;
}
}
printf("Case %d:
%d %d %d
",i,max,s,e);
if(i!=n)
printf("
");
}
return 0;
}