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

發布時間：2025/3/19 php 22 豆豆

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Triangular Sums

時間限制：3000 ms ?|? 內存限制：65535 KB 難度：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)]

輸入

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.

輸出

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.

樣例輸入

4 3 4 5 10

樣例輸出

1 3 45 2 4 105 3 5 210 4 10 2145

來源

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; }

一看題估計你會蒙了，但是一看代碼估計你會笑，其實此題不難，難的是理解不了題意。題意是給你一個數n，然后求出前n+1項和T（n+1），然后計算n*T(n+1)；輸出時注意格式就行了。

轉載于:https://www.cnblogs.com/wangyouxuan/p/3248825.html

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