HDOJ--1003

Max Sum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 192541    Accepted Submission(s): 44871

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

 

简单的贪心算法,，数组还是要开大一点。记得用两个变量记录起始和终止的值；

以下是AC代码：

#include<iostream>#include<cstring>using namespace std;int main(){        int t,num=0;    int n;    scanf("%d",&t);    getchar();    while(scanf("%d",&n)==1)    {        num++;        int a[100010];        int i,sum=0,start,end,k=1;        for(i=1;i<=n;i++)        cin>>a[i];        int max=-1001;        for(i=1;i<=n;i++)        {            sum+=a[i];            if(sum>max)            {                max=sum;                start=k;                end=i;            }            if(sum<0)            {                sum=0;                k=i+1;            }        }        cout<<"Case "<<num<<":"<<endl;        cout<<max<<" "<<start<<" "<<end<<endl;        if(num!=t)cout<<endl;            }    return 0; }