Problem A.Ant on a Chessboard
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Background
? One day, an ant called Alice came to an M*M chessboard. She wanted to go around all the grids. So she began to walk along the chessboard according to this way: (you can assume that her speed is one grid per second)

? At the first second, Alice was standing at (1,1). Firstly she went up for a grid, then a grid to the right, a grid downward. After that, she went a grid to the right, then two grids upward, and then two grids to the left…in a word, the path was like a snake.

? For example, her first 25 seconds went like this:

? ( the numbers in the grids stands for the time when she went into the grids)

?

25
?24
?23
?22
?21
?
10
?11
?12
?13
?20
?
9
?8
?7
?14
?19
?
2
?3
?6
?15
?18
?
1
?4
?5
?16
?17
?
5

4

3

2

1

?

1????????? 2????????? 3?????????? 4?????????? 5

At the 8th second , she was at (2,3), and at 20th second, she was at (5,4).

Your task is to decide where she was at a given time.

(you can assume that M is large enough)

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Input
? Input file will contain several lines, and each line contains a number N(1<=N<=2*10^9), which stands for the time. The file will be ended with a line that contains a number 0.

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Output
? For each input situation you should print a line with two numbers (x, y), the column and the row number, there must be only a space between them.

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Sample Input
8

20

25

0

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Sample Output
2 3

5 4

1 5

[cpp]
#include??
#include??
int main(void)?
{?
??? int n;?
??? while(scanf("%d",&n)&&n)?
??? {?
??????? int x=0,y=0;?
??????? int p=sqrt((double)n);?
??????? if(p*p==n&&p%2) {x=1;y=p;}?
??????? else if(p*p==n&&p%2==0){x=p;y=1;}?
??????? else if(p%2==0&&n!=p*p) {?
??????????? if(n-p-1<=p*p){x=p+1;y=n-p*p;}?
??????????? else{x=p-(n-1-p-p*p)+1;y=p+1;}?
??????? }?
??????? else if(p%2==1&&n!=p*p) {?
??????????? if(n-p-1<=p*p){x=n-p*p;y=p+1;}?
??????????? else{x=p+1;y=p-(n-1-p-p*p)+1;}?
??????? }?
??????? printf("%d %d\n",x,y);?
??? }?
??? return 0;?
}?

#include
#include
int main(void)
{
?int n;
?while(scanf("%d",&n)&&n)
?{
??int x=0,y=0;
??int p=sqrt((double)n);
??if(p*p==n&&p%2) {x=1;y=p;}
??else if(p*p==n&&p%2==0){x=p;y=1;}
??else if(p%2==0&&n!=p*p) {
???if(n-p-1<=p*p){x=p+1;y=n-p*p;}
???else{x=p-(n-1-p-p*p)+1;y=p+1;}
??}
??else if(p%2==1&&n!=p*p) {
???if(n-p-1<=p*p){x=n-p*p;y=p+1;}
???else{x=p+1;y=p-(n-1-p-p*p)+1;}
??}
??printf("%d %d\n",x,y);
?}
?return 0;
}

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