Max Sum
Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 52 Accepted Submission(s) : 12
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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
#include<iostream>
using namespace std;
int a[100005],b[100005],ans,ma;
int s1,s2,k,m,n,j,i,s,ll,kk,sum;
int main()
{
ll=0;
cin>>k;
for(s=1;s<=k;s++)
{
if(ll!=0)
cout<<endl;
else
ll=1;
ans=-100000000;
cin>>n;
memset(a,0,sizeof(a));
for(i=1;i<=n;i++)
scanf("%d",&b[i]);
a[1]=b[1];
sum=ma=b[1];
s1=1;s2=1;kk=1;
for(i=2;i<=n;i++)
{
if(ma<sum)
{ma=sum;s1=kk;s2=i-1;}
if(sum<0)
{ sum=0;kk=i;}
sum=sum+b[i];
}
if(ma<sum)
{ma=sum;s1=kk;s2=i-1;}
cout<<"Case "<<s<<":"<<endl;
printf("%d",ma);
cout<<' '<<s1<<' '<<s2<<endl;
}
return 0;
}