HDOJ——1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 82075    Accepted Submission(s): 18885

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

25 6 -1 5 4 -77 0 6 -1 1 -6 7 -5

Sample Output

Case 1:14 1 4Case 2:7 1 6

Author

Ignatius.L

for example:

9 1 -2 3 -7 4 8 4 9 10

i=1              1                          1

i=2              1 -2                      -1

i=3              3                          3

i=4              3 -7                      -4

i=5              4                          4

i=6              4 8                       12

i=7              4 8 4                    16

i=8              4 8 4 9                 25

i=9              4 8 4 9 10            35

 

截止位是负数也要加上。。。。这是为什么啊啊啊啊啊啊。。。先mark。。。容后再解决。。。

 

#include<iostream>using namespace std;int str[100003];int m[100003];int N;int ma;int start,end;int fstart,fend;void maxsum(){     for(int i=0;i<N;i++)    {        if(m[i]>=0)        {            m[i+1]=m[i]+str[i+1];            fend++;        }        if(m[i]<0)        {            m[i+1]=str[i+1];            fstart=fend=i+1;        }        if(m[i+1]>ma)        {            ma=m[i+1];            start=fstart;            end=fend;        }    }}void main(){    int T,i,j;    cin >> T ;    for(i=1;i<=T;i++)    {        fstart=fend=1;        ma=-1000000;        cin >> N;          memset (str ,-10000,N+2);         memset (m , -10000 , N+2);         for(j=1;j<=N;j++)        {            cin >> str[j];        }        maxsum();        cout << "Case " << i << ":" << endl << ma << " " << start << " " << end << endl ;        if(i<T)            cout << endl;    }}