HDU 1003 MAXSUM（最大子序列和）

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

 

题目主要是求连续最大子序列的和，并输出最大子序列的左边界和右边界。

注意输出的格式为每两组数据之间输出一个空行，最后一组数据没有。

 

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
#define INF 0x3f3f3f3f
int main()
{
    int t,n;
    int i,j,k=0;
    int Max,sum,x,y,l,d;
    scanf("%d",&t);
    while(t--)
    {
        Max=sum=-INF;
        scanf("%d",&n);
        for(i=1; i<=n; i++)
        {
            scanf("%d",&d);
            if(sum+d<d)
                sum=d,l=i;
            else
                sum+=d;
            if(Max<sum)
            {
                x=l;
                y=i;
                Max=sum;
            }
        }
        if(k)
            printf("
");
        printf("Case %d:
",++k);
        printf("%d %d %d
",Max,x,y);
    }
    return 0;
}