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Triangular Sums

2013-04-17 20:12:00.0 ACM之小试牛刀  
导读: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 ...。。。

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…nk * 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
434510
样例输出
1 3 452 4 1053 5 2104 10 2145

 

 

#include int t(int n){ return n*(n+1)/2;}int main(){ int N, n, sum, ans = 1; scanf("%d", &N); while(N--) {  sum = 0;  scanf("%d", &n);  for(int i = 1; i <= n; i++)  {   sum += i*t(i+1);  }  printf("%d %d %d\n", ans++, n, sum); } return 0;}


 

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