# HDU 1003 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
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

0 6 -1 1 -6 7 -5

1、二层for循环（会超时）

``````for i 0 to n
for j i to n
i，j表示左右端点``````

2、一层for循环

``````maxm=sum=arr[0]
for i 1 to n
if(sum<0)
sum+=arr[i]
else
sum=arr[i]
maxm=max(sum,maxm)``````

AC代码：

``````#include<cstdio>
#define maxn 200005
int main(){
int t,n,l=0;
int arr[maxn];
scanf("%d",&t);
while(t--){
if(l)
printf("\n");
scanf("%d",&n);
int sum,now,st,ed,ast,aed;
for(int i=0;i<n;++i)
scanf("%d",arr+i);
now=sum=arr[0];
st=ed=ast=aed=0;
for(int i=1;i<n;++i){
if(sum<0){
sum=arr[i];
st=ed=i;
}else{
sum+=arr[i];
ed=i;
}
if(sum>now){
now=sum;
ast=st;
aed=ed;
}

}
printf("Case %d:\n%d %d %d\n",++l,now,ast+1,aed+1);
}
}``````

HDU 1003 Max Sum【区间最大值】题解