Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 174588    Accepted Submission(s): 40639

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

题意：给定一个序列，求其子序列的最大和，还有最大和的子序列的起始位置和结束位置。

求和与end都没什么好说的。求start是亮点，一开始判断了很久，后来还是那个想法，对于序列或是字符串正着想想不出来就逆过来想，从左到右end容易求，那从右至左的话start就容易求了。

代码：

#include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #pragma warning(disable:4996) using namespace std;int value[100005];int dp[100005];int main(){	//freopen("input.txt","r",stdin);	//freopen("out.txt","w",stdout);	int Test,num,i,j,ans,start,end;	value[0]=0;	cin>>Test;	for(j=1;j<=Test;j++)	{		ans = -1005;		memset(dp,0,sizeof(dp));		cin>>num;		for(i=1;i<=num;i++)		{			cin>>value[i];			dp[i]=max(dp[i-1]+value[i],value[i]);			if(dp[i]>ans)			{				ans=dp[i];				end=i;			}		}		ans=-1005;		memset(dp,0,sizeof(dp));		for(i=num;i>=1;i--)		{			dp[i]=max(dp[i+1]+value[i],value[i]);			if(dp[i]>=ans)			{				ans=dp[i];				start=i;			}		}		cout<<"Case "<
         
          <<":"<
         
        
       
      
     
    

后来看discuss里其他人的思路，觉得从end开始往回倒，对每一个值都加起来，什么时候等于求出来的最大和了，什么时候就是start了，觉得这样做也很好。

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