Hdoj 1003.Max Sum 题解

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

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

Ignatius.L

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思路

最大连续子序列和问题，状态转移方程式：

\(f[i] = max(f[i-1]+a[i],a[i])\)

可以得出代码如下

代码

#include<bits/stdc++.h>
using namespace std;
const int INF = 1<<30;
int a[100001];
int main()
{
	int n;
	cin >> n;
	for(int q=1;q<=n;q++)
	{
		int len;
		cin >> len;

		int maxsum = -INF;
		int currentsum = 0;
		int l = 0,r = 0;
		int tmp = 1;
		for(int i=1;i<=len;i++)
		{
			cin >> a[i];
			if(currentsum >= 0)
				currentsum += a[i];
			else
			{
				currentsum = a[i];
				tmp = i;
			}
			if(currentsum > maxsum)
			{
				maxsum = currentsum;
				l = tmp;
				r = i;
			}
		}
		cout << "Case " << q << ":\n";
		cout << maxsum << " " << l << " " << r << endl;
		if(q!=n) cout << endl;
	}
	return 0;
}