HDU1003 Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 291913    Accepted Submission(s): 69228
 

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

用dp[i]表示以a[i]为end position的连续子序列的和的最大值。即dp[i]=max(a[i],dp[i-1]+a[i])。求出最大值Max后终点很容易得出，就是相应的dp[i]的下标。当最大值是负数时，起点和终点相等。当最大值非负时，起点为终点前连续的最后一次出现非负dp[i]的下标。

AC代码：

#include<iostream>
#include<cstring>
#include<algorithm>
#include<iomanip>
#include<cmath>
using namespace std;
int main()
{
    int T,dp[100001],a[100001],i,num,j=1;
    cin>>T;
    while(T--)
    {
        cin>>num;
        memset(dp,-1,sizeof(dp));
        memset(a,-1,sizeof(a));
        for(i=1;i<=num;i++)
        cin>>a[i];
        int i0=1,in=1,Max=a[1];
            for(i=1;i<=num;i++)
        {
           dp[i]=max(a[i],dp[i-1]+a[i]);
           if(dp[i]>Max)
           {
                   Max=dp[i];
                   i0=in=i;
           }
        }
            while(dp[i0]>=0) i0--;
            if(i0<in) i0++;
        cout<<"Case "<<j++<<':'<<endl;
        cout<<Max<<' '<<i0<<' '<<in<<endl;
        if(T>0) cout<<endl;//注意格式 
    } 
}