HDU 1003 Max Sum



Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 146529    Accepted Submission(s): 34242

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

#include <iostream>using namespace std;int main(){    int t,n,temp,pos1,pos2,max,now,x,i,j;    cin>>t;    for (i=1;i<=t;i++)    {        cin>>n>>temp;        now=max=temp;        pos1=pos2=x=1;//开始的时候默认最大的就是当前第一个        for (j=2;j<=n;j++)        {            cin>>temp;            if (now+temp<temp)//如果发现到现在的sum小于0果断舍弃当前的选择的序列                now=temp,x=j;            else                now+=temp;            if (now>max)//保存迄今为止最大的                max=now,pos1=x,pos2=j;        }        cout<<"Case "<<i<<":"<<endl<<max<<" "<<pos1<<" "<<pos2<<endl;        if (i!=t)            cout<<endl;    }    return 0;}