[ACM_动态规划] hdu1003 Max Sum [最大连续子串和]

 

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

 

???->>>>设某序列[x,j]的和大于等于0，则后面再来一个数肯定要接在上面；如果其小于0则后面来的数不要接在其后面，要把这个序列的开头设为当前输入的位置。

 

 1 #include <iostream>

 2 using namespace std;

 3 

 4 int main(){

 5     int t,n,temp,pos1,pos2,max,now,x,i,j;

 6     cin>>t;

 7     for (i=1;i<=t;i++){

 8         cin>>n>>temp;//输入n和第一个数

 9         now=max=temp;//令now和max都等于第一个数

10         pos1=pos2=x=1;//令初始位置为1

11         for(j=2;j<=n;j++){//循环输入2-n个

12             cin>>temp;

13             if (now+temp<temp)//如果now为负就从新的位置开始

14                 now=temp,x=j;//x保存当前开头位置

15             else//否则就在now后加上输入数据

16                 now+=temp;

17             if(now>max)//如果now>max就更新max

18                 max=now,pos1=x,pos2=j;

19         }

20         cout<<"Case "<<i<<":"<<endl<<max<<" "<<pos1<<" "<<pos2<<endl;

21         if (i!=t)cout<<endl;

22     }

23     return 0;

24 }