YT05-动态归划求解课后题目-1004—Max Sum -（6.21日-烟台大学ACM预备队解题报告）

Max Sum

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 10   Accepted Submission(s) : 9

Font: Times New Roman | Verdana | Georgia

Font Size: ← →

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 

一开始容易想到的都是暴力方法，一个个向后面走，只要和大于max的就记录，但是会超时

for (i=1;i<= m;i++)  
        {  
            int sum=0;  
            for (int j= i;j<m;j++)  
            {  
                sum+=data [j];  
                if (sum>max)  
                {  
                    max=sum;  
                    point1=i;  
                    point2=j;  
                }  
            }  
        }  

可以用动态规划，对算法进行改进，起始还是从头开始，一直往后搜索，当遇到和小于０的时候，就对sum重新赋值，并把记录起始的点变为和为0的后一位，和不小于0的，就继续搜索下去.

#include <iostream>
using namespace std;
int main()
{
     int T,n,i,max,num,sum;
     int jilu1,jilu2,qian,hou;
//定义jilu1、jilu2为运算中记录起始整数位置，qian、hou为最后结果
     cin>>T;
     for(int k=1;k<=T;k++)
     {
         cin>>n;
         sum=max=-1000;
         qian=hou=1;
         jilu1=jilu2=1;
         for(i=1;i<=n;i++)
         {
             cin>>num;
             if(sum>=0)		{jilu2++;sum+=num;}
else if(sum<0)
	 {sum=num;jilu1=i;jilu2=i;}
//当前面集合为负数时，不用将前面数字加入计算和，直接令sum = 新输入元素

	if(max<sum)         	
	{
	max=sum;
	qian=jilu1;
	hou=jilu2;
	}
 }
cout<<max<<‘ '<<qian<<‘ '<<hou<<endl;
	if(k!=T)
	cout<<endl;	
     }
     return 0;
}