The following program is my ACcepted code for UVA-591.
It's a for everybody to learn and discuss.
If there is any mistake or comment, please let me know. :D
此乃UAV 591 的AC code!
歡迎一同討論學習,如有錯誤與任何建議請留言 : )
//This program is for UVA 913 Joana and the Odd Numbers
//http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=11&page=show_problem&problem=854
/* Do some math!
calculate the height(h): (1+n)/2
then calculate the number of the odd number: x = (1+n)*h / 2 = ( (1+n)^2 ) / 4
then calculate the last number: y = x*2 - 1
the final answer is the last three number so we have: y*3-6
*/
#include<stdio.h>
using namespace std;
int main()
{
long long int n, ans;
while(scanf("%lld", &n)!=EOF)
{
ans = (1 + n)*( 1 + n) / 4;
ans = ans * 2 - 1;
ans = ans * 3 -6;
printf("%lld\n", ans);
}
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。 : )
It's a for everybody to learn and discuss.
If there is any mistake or comment, please let me know. :D
此乃UAV 591 的AC code!
歡迎一同討論學習,如有錯誤與任何建議請留言 : )
//This program is for UVA 913 Joana and the Odd Numbers
//http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=11&page=show_problem&problem=854
/* Do some math!
calculate the height(h): (1+n)/2
then calculate the number of the odd number: x = (1+n)*h / 2 = ( (1+n)^2 ) / 4
then calculate the last number: y = x*2 - 1
the final answer is the last three number so we have: y*3-6
*/
#include<stdio.h>
using namespace std;
int main()
{
long long int n, ans;
while(scanf("%lld", &n)!=EOF)
{
ans = (1 + n)*( 1 + n) / 4;
ans = ans * 2 - 1;
ans = ans * 3 -6;
printf("%lld\n", ans);
}
return 0;
}
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。 : )
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