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):
X
X X
X X X
X X X XWrite 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
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1 3 45 2 4 105 3 5 210 4 10 2145
- 来源
- Greater New York 2006
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#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。