HDOJ--1003--Max Sum【最大子段和】

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

 

Author

Ignatius.L

 

最大子段和的入门题。。。

#include
#include
#include
#include
#include
#include
using namespace std;
int n,s,e;
int a[100005];

int maxsub(int a[])
{
    s=e=1;
    int sum,j,b,bs,be;
    sum=b=a[0];
    for(j=0;j=0&&j!=0)
        {
            b=b+a[j];
            be=j+1;
        }
        else
        {
            b=a[j];
            bs=j+1;
            be=j+1;
        }
        if(b>sum)
        {
            sum=b;
            s=bs;
            e=be;
        }
    }
    return sum;
}

int main()
{
    int t,i,k=1,sum;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=0;i