Light oj 1008

Description

Fibsieve had a fantabulous (yes, it's an actual word) birthday party this year. He had so many gifts that he was actually thinking of not having a party next year.

Among these gifts there was an N x N glass chessboard that had a light in each of its cells. When the board was turned on a distinct cell would light up every second, and then go dark.

The cells would light up in the sequence shown in the diagram. Each cell is marked with the second in which it would light up.

(The numbers in the grids stand for the time when the corresponding cell lights up)

In the first second the light at cell (1, 1) would be on. And in the 5th second the cell (3, 1) would be on. Now, Fibsieve is trying to predict which cell will light up at a certain time (given in seconds). Assume that N is large enough.

Input

Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case will contain an integer S (1 ≤ S ≤ 10¹⁵) which stands for the time.

Output

For each case you have to print the case number and two numbers (x, y), the column and the row number.

Sample Input

3

8

20

25

Sample Output

Case 1: 2 3

Case 2: 5 4

Case 3: 1 5

题意：在如上图所示的图中，例如1的坐标为(1,1)， 5的坐标为(3,1)。给出任意n，问n的坐标。

题解：规律题。观察图中填数的走向（即不同的颜色标示），发现每个走向的数的个数为1  3  5   7   9.........

而每次走向中每个数n开平方取整的值cnt相同。而奇数次走向和偶数次走向完全相反。归纳寻找，cnt为偶数时，当cnt*cnt-n<cnt时，x=cnt，y=cnt*cnt-n+1。否则y=cnt，x=n-cnt*cnt+2*cnt-1。 当cnt为奇数时，则交换x，y的值。（对称）。

用long long 的情况下，在编译器上测试数据的时候得不到正确结果，只要感觉对就可以提交；

代码：

#include<stdio.h>#include<string.h>#include<math.h>int main(){int b;int mm=1;long long n,x,y;scanf("%d",&b);while(b--){scanf("%lld",&n);long long t=sqrt(n);if(t*t==n){if(t%2==0){    x=t;    y=1;}else{x=1;y=t;}}else{t=t+1;if(t%2==0){if(t*t-n<t)    {   x=t;   y=t*t-n+1;    }    else    {   y=t;   x=n-(t-1)*(t-1);    }}else {if(t*t-n<t){x=t*t-n+1;y=t;}else{x=t;y=n-(t-1)*(t-1);}}}printf("Case %d: %lld %lld\n",mm++,x,y);}return 0;}

我自己的思路，就因为用long long在编译器上测试数据通不过，所以没有提交，在比赛中少做了两道题，还是很重要的比赛，全坑在long long 上，呜呜。。。。。。

代码2：

#include<stdio.h>#include<string.h>#include<math.h>int main(){int b;int mm=1;long long n,x,y;scanf("%d",&b);while(b--){scanf("%lld",&n);long long t=sqrt(n);if(t*t==n){if(t%2==0){    x=t;    y=1;}else{x=1;y=t;}}else{t=t+1;long long w=n-(t-1)*(t-1);if(t%2==0){if(w<=t)    {   x=w;   y=t;    }    else    {   x=t;   y=2*t-w;    }}else if(t%2!=0){if(w<=t){x=t;y=w;}else{x=2*t-w;y=t;}}}printf("Case %d: %lld %lld\n",mm++,x,y);}return 0;}