南阳oj入门题-Triangular Sums

最新推荐文章于 2025-12-03 10:19:14 发布

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本文介绍了一种计算加权三角形数之和的算法，即W(n)=Σ[k=1...n;k*T(k+1)]，其中T(n)为第n个三角形数。文章提供了完整的C++代码实现，通过输入三角形的一边长度n，输出对应的加权三角形数之和。

/**
Triangular Sums
时间限制：3000 ms  |  内存限制：65535 KB
难度：2
描述
The nth Triangular number, T(n) = 1 + … + n, is the sum of the first n integers. It is the number of points in a triangular array with n points on side. For example T(4):

X
X X
X X X
X X X X
Write a program to compute the weighted sum of triangular numbers:

W(n) = SUM[k = 1…n; k * T(k + 1)]

输入
The first line of input contains a single integer N, (1 ≤ N ≤ 1000) which is the number of datasets that follow.

Each dataset consists of a single line of input containing a single integer n, (1 ≤ n ≤300), which is the number of points on a side of the triangle.
输出
For each dataset, output on a single line the dataset number (1 through N), a blank, the value of n for the dataset, a blank, and the weighted sum ,W(n), of triangular numbers for n.
样例输入
4
3
4
5
10
样例输出
1 3 45
2 4 105
3 5 210
4 10 2145
*/
#include<iostream>
using namespace std;
int main()
{
	int N;
	cin>>N;
	while(N--)
	{
		int i,j=1,k;
		int sum=0;
		cin>>k;
		for(i=1;i<=k;i++)
			sum+=(i*(i+1)*(i+2))/2;
		cout<<j++<<" "<<k<<" "<<sum<<endl;			
	}
	return 0;
}