HDU--1003 Max Sum（最大连续子序列和）

Posted 2020-10-01

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

 

 

Recommend

We have carefully selected several similar problems for you:  1176 1069 2084 1058 1203 

代码：

 1 #include<cstdio>
 2 int a[100000+10];
 3 int main()
 4 {
 5     int t,n,k=1;
 6     scanf("%d",&t);
 7     while(t--)
 8     {
 9         scanf("%d",&n);
10         for(int i=1;i<=n;i++)
11             scanf("%d",&a[i]);
12         int max=-1001,st,endd,sum=0,st1=1; //注意整数的范围
13         for(int i=1;i<=n;i++)     
14         {
15             sum+=a[i];
16             if(sum>max)
17             {
18                 max=sum;
19                 st=st1;
20                 endd=i;    
21             }
22             if(sum<0)
23             {
24                 sum=0;
25                 st1=i+1;    //st1是临时的开始点，如果后面的sum<0，这个开始点也就不会计入
26             }    
27         }
28         printf("Case %d:\n%d %d %d\n",k++,max,st,endd);
29         if(t!=0)
30         {
31             printf("\n");
32         }
33         
34     }
35     return 0;
36 }