nyoj 122-Triangular Sums (数学之读懂求和公式的迭代)

122-Triangular Sums

内存限制:64MB
时间限制:3000ms
特判: No

通过数:5
提交数:7
难度:2

题目描述:

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)]

输入描述:

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.

输出描述:

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:

复制

4
3
4
5
10

样例输出:

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

C/C++ AC：

 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #include <cstdio>
 #include <cmath>
 #include <stack>
 #include <set>
 #include <map>
 #include <queue>
 #include <climits>

 using namespace std;
 const int MY_MAX = ;
 int T[MY_MAX] = {}, W[MY_MAX] = {}, n;

 void cal_W()
 {
     T[] = , T[] = ;
     for (int i = ; i < MY_MAX - ; ++ i)
     {
         W[i] = W[i - ] + i * T[i + ];
         T[i + ] = T[i + ] + i + ;
     }
 }

 int main()
 {
     cal_W();
     cin >>n;
     for (int i = ; i <= n; ++ i)
     {
         int temp;
         scanf("%d", &temp);
         printf("%d %d %d\n", i, temp, W[temp]);
     }
 }