HDU_oj_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

 分析：

此题为连续子序列求最大和问题，主要思路有①枚举②累积遍历③动态规划④分治

方法比较多，将会持续更新所有算法

注意点：

①每种case中间的空行

②注意算法复杂度，枚举肯定行不通

可以阅读以下博文，详细了解各种解法：

https://www.cnblogs.com/TWS-YIFEI/p/5590532.html

http://blog.csdn.net/hcbbt/article/details/10454947

//累积遍历算法 
/*
算法复杂度 O(n)

*/ 

#include<iostream>
#define MAX 100100

int main()
{
    int num[MAX+2]={0};
    int t,n;
    int sum,max;
    int m,kk,ll;
    
    scanf("%d",&t);
    for(int i=1;i<=t;i++)
    {
        scanf("%d",&n);
        scanf("%d",&num[0]);
        sum=max=num[0];
        m=kk=ll=0;
        
        for(int j=1;j<n;j++)
        {
            scanf("%d",&num[j]);
            
            if(sum<0)
            {
                sum=num[j];
                m=j;
            }
            else
                sum+=num[j];
            
            if(sum>max)
            {
                max=sum;
                kk=m;
                ll=j;
            }    
        }
        if(i!=1)
        printf("
");
        printf("Case %d:
",i);
        printf("%d %d %d
",max,kk+1,ll+1);
    }
    return 0;
}