Max Sum
Time Limit : 2000/1000ms (Java/Other) Memory Limit :65536/32768K (Java/Other)
Total Submission(s) : 6 Accepted Submission(s) : 0
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Problem Description
Given asequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of asub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is6 + (-1) + 5 + 4 = 14.
Input
The firstline of the input contains an integer T(1<=T<=20) which means the numberof test cases. Then T lines follow, each line starts with a numberN(1<=N<=100000), then N integers followed(all the integers are between-1000 and 1000).
Output
For each testcase, you should output two lines. The first line is "Case #:", #means the number of the test case. The second line contains three integers, theMax Sum in the sequence, the start position of the sub-sequence, the endposition of the sub-sequence. If there are more than one result, output thefirst 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
Author
//*************************
//* 任务介绍：求最大的和 *
//* 作者：何香 *
//* 完成时间：2013.10.21 *
//*************************
#include <iostream>
using namespace std;
int main ()
{
int T,N,temp,max,sum,end,start,x;
/*
T 测试数据组数
N 每组数据的长度
temp 当前取的数据
start 最后MAX SUM的起始位置
end 最后MAX SUM的结束位置
max 当前得到的MAX SUM
sum 在读入数据时，能够达到的最大和
x 记录最大和的起始位置，因为不知道跟之前的max值的大小比，所以先存起来
*/
int i,j;
cin>>T;
for(j=1;j<=T;++j)
{
cin>>N>>temp;//先输入第一个数据
sum=max=temp;
end=1,start=1,x=1;
for(i=2;i<=N;++i)//再从第二个数据开始输入
{
cin>>temp;
if(sum+temp<temp)//如果新输入的数据加上以前的变小了，则删除，并从现在这一项开始
{
sum=temp;
x=i;//开始的序号
}
else
sum+=temp;
if(sum>max)
{
max=sum;
//start=x;
end=i;//结束的序号
}
}
cout<<"Case "<<j<<":"<<endl<<max<<" "<<start<<" "<<end<<endl;
if(j!=T)
{
cout<<endl;
}
}
return 0;
}