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.
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).
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.
2
5 6 -1 5 4 -7
7 0 6 -1 1 -6 7 -5
Case 1:
14 1 4
Case 2:
7 1 6
//刚开始做这道题时理解错了，之后参考了别人的代码知道自己错那了，，之后还是出错，忘了考虑一种情况
如果是2 -8 9 18 -69 28输出结果应该为 28 6 6 但我的代码结果为28 5 6
#include<iostream>
#include<stdio.h>
int a[100000],min=-10000000;
using namespace std;
int main()
{
int t,n,max;
int i,j,k,sum,s,e;
cin>>t;
for(i=1;i<=t;i++)
{
cin>>n;
max=min;
for(j=1;j<=n;j++)
cin>>a[j];
sum=0;
k=1;
for(j=1;j<=n;j++)
{
sum+=a[j];
if(sum>max)
{
max=sum;
s=k;
e=j;
}
if(sum<0)
{
sum=0;
k=j+1;
}
}
printf("Case %d:\n",i);
cout<<max<<" "<<s<<" "<<e<<endl;
if(i!=t)
cout<<endl;
}
return 0;
}
