Triangular Sums
| Time Limit: 1000MS |
|
Memory Limit: 65536K |
| Total Submissions: 6371 |
|
Accepted: 4529 |
Description
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…n; k * T(k + 1)]
Input
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.
Output
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.
Sample Input
4
3
4
5
10
Sample Output
1 3 45
2 4 105
3 5 210
4 10 2145
Source
Greater New York 2006 找规律的一道题，其实题目已经说的很明白了。 先打表再输出，开始怕时限，所以一直改不对，干脆分开算。 代码：
#include
#include
#include
using namespace std; int f[305],T[305],sum[305]; int main() { int i,j,n,m; T[1]=1;f[1]=1; for(i=2;i<305;i++) T[i]=i+T[i-1]; for(i=1;i<304;i++) f[i]=i*T[i+1]; sum[1]=f[1]; for(i=2;i<305;i++) sum[i]=sum[i-1]+f[i]; while(scanf("%d",&n)!=EOF&&n) { for(i=1;i<=n;i++) { scanf("%d",&m); printf("%d %d %d\n",i,m,sum[m]); } } return 0; }