Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

再谈动态规划！

具体代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int dp[100001];
int a[100001];
int s[100001];
int main()
{
    int T,k,count=0;
    cin>>T;
    while(T--)
    {
        cin>>k;
        count++;
        memset(a,0,sizeof(a));
        memset(dp,0,sizeof(dp));
        s[0]=s[1]=1;
        for(int i=1;i<=k;i++)
        {
            cin>>a[i];
        }
        for(int i=1;i<=k;i++)
        {
            if(dp[i-1]+a[i]>=a[i])
            {
                dp[i]=dp[i-1]+a[i];
                s[i]=s[i-1];
            }
            else
            {
                dp[i]=a[i];
                s[i]=i;
            }
        }
        int start=1,end=1;
        int max=dp[1];
        for(int i=2;i<=k;i++)
        {
            if(max<dp[i])
            {
                max=dp[i];
                start=s[i];
                end=i;
            }
        }
        cout<<"Case "<<count<<":"<<endl;
        cout<<max<<" "<<start<<" "<<end<<endl;
        if(T!=0)cout<<endl;
    }
    return 0;
}

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