poj--3086

escription

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…nk * 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

具体代码:

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
using namespace std;

int main()
{
int t, n;
scanf("%d", &t);
for (int i = 0; i < t; i++)
{
scanf("%d", &n);
int w = 0, x = 1;
for (int j = 1; j <= n; j++)
{
x += j + 1;
w += j * x;
}
printf("%d %d %d\n", i + 1, n, w);
}
return 0;
}

poj--3086

时间: 2024-11-01 23:09:12

poj--3086的相关文章

POJ 3086 Triangular Sums

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 po

POJ - 3186 Treats for the Cows (区间DP)

题目链接:http://poj.org/problem?id=3186 题意:给定一组序列,取n次,每次可以取序列最前面的数或最后面的数,第n次出来就乘n,然后求和的最大值. 题解:用dp[i][j]表示i~j区间和的最大值,然后根据这个状态可以从删前和删后转移过来,推出状态转移方程: dp[i][j]=max(dp[i+1][j]+value[i]*k,dp[i][j-1]+value[j]*k) 1 #include <iostream> 2 #include <algorithm&

POJ 2533 - Longest Ordered Subsequence(最长上升子序列) 题解

此文为博主原创题解,转载时请通知博主,并把原文链接放在正文醒目位置. 题目链接:http://poj.org/problem?id=2533 Description A numeric sequence of ai is ordered if a1 < a2 < ... < aN. Let the subsequence of the given numeric sequence (a1, a2, ..., aN) be any sequence (ai1, ai2, ..., aiK)

POJ——T2271 Guardian of Decency

http://poj.org/problem?id=2771 Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 5932   Accepted: 2463 Description Frank N. Stein is a very conservative high-school teacher. He wants to take some of his students on an excursion, but he is

POJ——T2446 Chessboard

http://poj.org/problem?id=2446 Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 18560   Accepted: 5857 Description Alice and Bob often play games on chessboard. One day, Alice draws a board with size M * N. She wants Bob to use a lot of c

poj 1088 滑雪 DP(dfs的记忆化搜索)

题目地址:http://poj.org/problem?id=1088 题目大意:给你一个m*n的矩阵 如果其中一个点高于另一个点 那么就可以从高点向下滑 直到没有可以下滑的时候 就得到一条下滑路径 求最大的下滑路径 分析:因为只能从高峰滑到低峰,无后效性,所以每个点都可以找到自己的最长下滑距离(只与自己高度有关).记忆每个点的最长下滑距离,当有另一个点的下滑路径遇到这个点的时候,直接加上这个点的最长下滑距离. dp递推式是,dp[x][y] = max(dp[x][y],dp[x+1][y]+

POJ 1385 计算几何 多边形重心

链接: http://poj.org/problem?id=1385 题意: 给你一个多边形,求它的重心 题解: 模板题,但是不知道为啥我的结果输出的确是-0.00 -0.00 所以我又写了个 if (ans.x == 0) ans.x = 0 感觉好傻逼 代码: 1 #include <map> 2 #include <set> 3 #include <cmath> 4 #include <queue> 5 #include <stack> 6

POJ 1741 Tree(树的点分治,入门题)

Tree Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 21357   Accepted: 7006 Description Give a tree with n vertices,each edge has a length(positive integer less than 1001).Define dist(u,v)=The min distance between node u and v.Give an in

poj 1655 树的重心

Balancing Act Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 13178   Accepted: 5565 Description Consider a tree T with N (1 <= N <= 20,000) nodes numbered 1...N. Deleting any node from the tree yields a forest: a collection of one or m

POJ 2479 Maximum sum

http://poj.org/problem?id=2479 题意: 给出一个整数串,求连续子串1和连续子串2,不相交并且串1加串2的和最大. 思路: 其实就是求最大连续和,题意要求就是求两段最大连续和.我们可以从左边和右边分别求最大连续和,代码中的dp_l[i]就是1~i的最大连续和,dp_r[i]则是i~n的最大连续和. 1 #include<iostream> 2 #include<string> 3 #include<cstring> 4 #include<