nyoj-122

Triangular Sums

时间限制：3000 ms  |  内存限制：65535 KB

难度：2

描述

The n^th 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

我的代码：

 
#include<iostream>
#include<stdio.h>
#include<cmath>
using namespace std;
int main()
{

	int b[1001]={0};
	int n,m=0,w;
	int p=1;
	int T=0;
	cin>>w;

	while(w--){
	cin>>n;

	       for(int i=1;i<=n;i++)
		   {       
		      for(int j=1;j<=i+1;j++)
		   
                     b[j]=b[j-1]+j;
		      m= m+i*b[i+1];
		
		   }
	      printf("%d %d %d\n",p++,n,m);
           m=0;
	
	}
	return 0;
}
        

解题关键：

W(n) = SUM[k = 1…n; k * T(k + 1)]；

T(n) = 1 + … + n；

两个公式即可。

反思。。。没有一次A过，是因为n的取值范围和N 的取值范围混淆了，看题要仔细呀！亲，不要把时间放在检查这种马虎问题上，太浪费啦！！！面壁思过。。。