hdu-1003-Max Sum

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 157118    Accepted Submission(s): 36725

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

 

Author

Ignatius.L

 

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#include<iostream>using namespace std;int main() {    int n,count=0;    cin>>n;    while(n--) {        int a[100001]= {0},m;        cin>>m;        for(int i=0; i<m; i++)            cin>>a[i];        int max=-1000,start=0,end=1,sum=0,newstart=0;        for(int i=0; i<m; i++) {            sum+=a[i];            if(sum>max) {                max=sum;                start=newstart+1;                end=i+1;            }            if(sum<0) {//这里要让i往后加，他这肯定会产生正数                newstart=i+1;                sum=0;            }        }        count++;        cout<<"Case "<<count<<":"<<endl;        cout<<max<<" "<<start<<" "<<end<<endl;        if(n)cout<<endl;//不能输出多余的换行，否则就是PE    }    return 0;}