HDU--Max Sum

小咪咪 2022-08-26 05:52 5阅读 0赞

# Max Sum #

**Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 131345    Accepted Submission(s): 30431**

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
      
       
      
       
       
        解析：
       
        对于整个序列a[n]来说，它的所有子序列有很多很多，但是可以将它们归类。 
       
        注意，是以**结尾的子序列，其中肯定是要包含**的了 
       
        
       
        以a[0]结尾的子序列只有a[0] 
       
        以a[1]结尾的子序列有 a[0]a[1]和a[1] 
       
        以a[2]结尾的子序列有 a[0]a[1]a[2] / a[1]a[2] / a[2] 
       
        …… 
       
        以a[i]结尾的子序列有a[0]a[1]……a[i-2]a[i-1]a[i]  / a[1]a[2]……a[i-2]a[i-1]a[i] /  a[2]a[3]……a[i-2]a[i-1]a[i] / …… /  a[i-1]a[i] / a[i] 
       
        
       
        所有以a[0] ~a[n]结尾的子序列分组构成了整个序列的所有子序列。 
       
        
       
        这样，我们只需求以a[0]~a[n]结尾的这些分组的子序列中的每一分组的最大子序列和。然后从n个分组最大子序列和中选出整个序列的最大子序列和。 
       
        
       
        观察可以发现，0，1，2，……，n结尾的分组中， 
       
        maxsum a[0] = a[0] 
       
        maxsum a[1] = max( a[0] + a[1] ，a[1])  =  max( maxsum a[0] + a[1] ，a[1])  
       
        maxsum a[2] = max( max ( a[0] + a[1] + a[2]，a[1] + a[2] )，a[2])   
       
        = max(  max( a[0] + a[1] ，a[1]) + a[2] ， a[2])  
       
        = max(  maxsum a[1] + a[2] ， a[2]) 
       
        …… 
       
        依此类推，可以得出通用的式子。 
       
        maxsum a[i] = max（ maxsum a[i-1] + a[i]，a[i]) 
       
        #include <iostream>  
    using namespace std;  
    int main()  
    {  
        int j,i,k,n,m,t;  
        int a;  //不需要数组，只需要一个输入变量
        scanf("%d",&t);  
        for (j=1;j<=t;j++)  
        {  
            scanf("%d",&n);  
            int sum=0,maxsum=-1001,first =0, last = 0, temp = 1;  
            for (i=0;i<n;i++)  
            {  
                scanf("%d",&a); 
                sum += a;  
                if (sum > maxsum)  
                {  
                    maxsum = sum;first = temp;last = i+1;  
                }  
                if (sum < 0)  
                {  
                    sum = 0;temp = i+2;  
                }  
            }  
      //注意格式，我就因为将冒号写到了数的前边而wrong answer，郁闷半天才发现……
            printf("Case %d:\n%d %d %d\n",j,maxsum,first,last);  
            if (j!=t)  
            {  
                printf("\n");  
            }  
        }  
        return 0;  
    }