Let's consider a triangle of numbers in which a number appears in the first line, two numbers appear in the second line, three in the third line, etc. Develop a program which will compute the largest of the sums of numbers that appear on the paths starting
from the top towards the base, so that:

• on each path the next number is located on the row below, more precisely either directly below or below and one place to the right;
• the number of rows is strictly positive, but less than 100
• all numbers are positive integers between O and 99.

Input

In the first line integer n - the number of test cases (equal to about 1000).

Then n test cases follow. Each test case starts with the number of lines which is followed by their content.

Output

For each test case write the determined value in a separate line.

Example

Input:

2

3

1

2 1

1 2 3

4

1

1 2

4 1 2

2 3 1 1 

Output:

5

9

使用动态规划法解决本题。

和一题leetcode题一样，从底往上查找路径。

#include <vector>

#include <string>

#include <algorithm>

#include <stdio.h>

#include <iostream>

using namespace std;

int triPath(vector<vector<int> > &tri)

{

	if (tri.empty()) return 0;

	vector<int> path(tri.back());

	for (int i = (int)tri.size() - 2; i >= 0 ; i--)//unsigned做减法会溢出！！！

	{

		for (int j = 0; j < (int)tri[i].size(); j++)

		{

			path[j] = max(path[j], path[j+1]) + tri[i][j];

		}

	}

	return path.front();

}

int SumsinATriangle()

{

	int T, n;

	scanf("%d", &T);

	while (T--)

	{

		scanf("%d", &n);

		vector<vector<int> > tri;

		for (int i = 1; i <= n; i++)

		{

			vector<int> tmp(i);

			for (int j = 0; j < i; j++)

			{

				scanf("%d", &tmp[j]);

			}

			tri.push_back(tmp);

		}

		printf("%d\n", triPath(tri));

	}

	return 0;

}