UVa OJ - 12149 Feynman

The following program is my ACcepted code for UVA-12149 .
It's a for everybody to learn and discuss.
If there is any mistake or comment, please let me know.  :D

此乃UVA 12149 的AC code!

歡迎一同討論學習，如有錯誤與任何建議請留言 : )

點這裡看題目 Click here to see this Problem!

一個簡單的數學問題
當邊為 1 時，會有  n  *  n   squares.
當邊為 2 時，會有(n-1)*(n-1) squares.
.
.
直到...
當邊為 n 時，會有  1  *  1   squares.
全部加總會得到答案!

A simple math problem.
When the edge = 1, there're  n  *  n   squares.
When the edge = 2, there're(n-1)*(n-1) squares.
.
.
Until..
When the edge = n, there're  1  *  1   squares.
Then add them to get the answer.


//This program is for UVA 12149 Feynman
//題目來源 Problem link: http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=243&page=show_problem&problem=3301

#include<stdio.h>
using namespace std;

int calSquar(int n)
{
    int sum = 0;
 
    for(int i = 1; i <= n; i++)
        sum = sum + (i * i);
     
    return sum;  
}

int main()
{
    int n;
 
    while(scanf("%d", &n) && n)
        printf("%d\n", calSquar(n));
    return 0;
}

Please feel free to use it after adding this blog as an reference. (http://autekroy.blogspot.tw) If there is any mistake or comment, please let me know. :D 

歡迎使用與分享任何內容，但請記得標示此部落格為出處。(http://autekroy.blogspot.tw/) 如果有發現任何的錯誤與建議請留言或跟我連絡。 : )