Triangular Sums http://acm.nyist.net/JudgeOnline/problem.php?pid=122

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):

[align=center]X
X X
X X X
X X X X[/align]
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

来源Greater New York 2006

#include<stdio.h>
int main()
{
int i,n;
scanf("%d",&n);
for(i=1;i<=n;i++)
{
int k,j,m;
long int toal=0,sum=1;
scanf("%d",&m);
for(j=1;j<=m;j++)
{
sum+=j+1;
toal=toal+j*sum;
}
printf("%d %d %ld\n",i,m,toal);
}
return 0;
}

一看题估计你会蒙了，但是一看代码估计你会笑，其实此题不难，难的是理解不了题意。题意是给你一个数n，然后求出前n+1项和T（n+1），然后计算n*T(n+1)；输出时注意格式就行了。