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])

//#define LOCAL

#include<cstdio>

const int INF=-1000000;

int sum,sum_max,begin,end,temp,T,N,conn;

void solve()

{

    scanf("%d",&N);

    int a;

    sum_max=INF,sum=0,temp=1,begin=1;

    for(int i=0;i<N;i++)

    {

        scanf("%d",&a);

        sum=sum+a;

        if(sum_max<sum)

        {

            sum_max=sum;

            begin=temp;

            end=i+1;

        }

        if(sum<0)

        {

            sum=0;

            temp=i+2;

        }

    }

    printf("Case %d:\n%d %d %d\n",(conn-T),sum_max,begin,end);

    if(T!=0)

    {

        printf("\n");

    }

}

int main()

{

#ifdef LOCAL

    freopen("1003.in","r",stdin);

    freopen("1003.out","w",stdout);

#endif

    scanf("%d",&T);

    conn=T;

    while(T--)

    {

        solve();

    }

    return 0;

}

Max Sum

Time
Limit: 2000/1000 MS (Java/Others)    Memory Limit:
65536/32768 K (Java/Others)
Total Submission(s):
136354    Accepted Submission(s):
31568

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

sum max(hdu 1003)