HDU 1003 B - Max Sum

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

 

 

 

 

以下是变量说明：t         测试数据组数n         每组数据的长度temp         当前取的数据pos1   最后MAX SUM的起始位置pos2         最后MAX SUM的结束位置max 当前得到的MAX SUMnow 在读入数据时，能够达到的最大和x 记录最大和的起始位置，因为不知道跟之前的max值的大小比，所以先存起来下面模拟过程：1.首先，读取第一个数据，令now和max等于第一个数据，初始化pos1,pos2,x位置2.然后，读入第二个数据，判断①. 若是now+temp<temp，表示当前读入的数据比之前存储的加上当前的还大，说明可以在当前另外开始记录，更新now=temp②. 反之，则表示之前的数据和在增大，更新now=now+temp3.之后，把now跟max做比较，更新或者不更新max的值，记录起始、末了位置4.循环2~3步骤，直至读取数据完毕。

 

 

 

#include <iostream>  

 

using namespace std;  

 

  

 

int main()  

 

{  

 

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

 

    cin>>t;  

 

    for (i=1;i<=t;i++)  

 

    {  

 

        cin>>n>>temp;  

 

        now=max=temp;  

 

        pos1=pos2=x=1;  

 

        for (j=2;j<=n;j++)  

 

        {  

 

            cin>>temp;  

 

            if (now+temp<temp)  

 

                now=temp,x=j;  

 

            else  

 

                now+=temp;  

 

            if (now>max)  

 

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

 

        }  

 

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

 

        if (i!=t)  

 

            cout<<endl;  

 

    }  

 

    return 0;  

 

}  

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

#include <iostream>

#include <stdio.h>

using namespace std;

int main()

{

    int t,l;

    scanf("%d",&t);

    for(l=1;l<=t;l++)

    {

        int n,temp,p1,p2,sum,now,k,i;

        scanf("%d%d",&n,&temp);

        now=sum=temp;

        p1=p2=k=1;

        for(i=2;i<=n;i++)

        {

            scanf("%d",&temp);

            if(now+temp<temp)

            {

                now=temp;

                k=i;

            }

            else

                now+=temp;

            if(now>sum)

            {

                sum=now;

                p1=k;

                p2=i;

            }

        }

        printf("Case %d:\n",l);

        printf("%d %d %d\n",sum,p1,p2);

        if(l!=t)

            printf("\n");

    }

    return 0;

}

 

 

 

 

 

#include <iostream>

#include <stdio.h>

using namespace std;

int main()

{

    int t,w=1;

    scanf("%d",&t);

    while(t--)

    {

        int n,temp,p1,p2,sum,now,k,i;

        scanf("%d%d",&n,&temp);

        now=sum=temp;//记录和 与 最大值

        p1=p2=k=1;

        for(i=2;i<=n;i++)

        {

            scanf("%d",&temp);

            if(now+temp<temp)

            {

                now=temp;//重开起点

                k=i;

            }

            else

                now+=temp;

            if(now>sum)

                sum=now,p1=k,p2=i;//更新最大值 记录开始与到最大值的结束位置

        }

        printf("Case %d:\n",w++);

        printf("%d %d %d\n",sum,p1,p2);

        if(t!=0)

            printf("\n");

    }

    return 0;

}