hdu1003max sum（尺取法）

题目链接https://vjudge.net/problem/HDU-1003

题目如下

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

这道题用尺取法代码比较简洁，也不会超时

先从第一个数开始数计算，当子序列和小于零时，放弃这个子序列，从下一个数开始计算，计算下一个子序列

代码如下

#include<cstdio>
#include<iostream>
using namespace std;
int main()
{
	int t,n,maxsum,maxleft,maxright,thissum;
	int kase=0,thisleft;
	cin>>t;
	while(t--)
	{
		cin>>n;
		cin>>maxsum;
		thisleft=maxleft=maxright=1;
		thissum=maxsum;
		if(maxsum<0)
		{
			thissum=0;
			thisleft=2;
		}
		for(int i=2;i<=n;i++)
		{
			int tem;
			cin>>tem;
			thissum+=tem;
			if(thissum>maxsum)
			{
				maxsum=thissum;
				maxleft=thisleft;
				maxright=i;
			}
			if(thissum<0)
			{
				thissum=0;
				thisleft=i+1;
			}
		}
		cout<<"Case "<<++kase<<":"<<endl;
		cout<<maxsum<<" "<<maxleft<<" "<<maxright<<endl;
		if(t)printf("\n");
	}
	return 0;
}