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 = 310;
int T[MY_MAX] = {0}, W[MY_MAX] = {0}, n;

void cal_W()
{
    T[1] = 1, T[2] = 3;
    for (int i = 1; i < MY_MAX - 5; ++ i)
    {
        W[i] = W[i - 1] + i * T[i + 1];
        T[i + 2] = T[i + 1] + i + 2;
    }
}

int main()
{
    cal_W();
    cin >>n;
    for (int i = 1; i <= n; ++ i)
    {
        int temp;
        scanf("%d", &temp);
        printf("%d %d %d
", i, temp, W[temp]);
    }
}