10161 - Ant on a Chessboard
Problem A.Ant on a Chessboard
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)
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.
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.
Sample Input
8
20
25
0
Sample Output
2 3
5 4
1 5
[cpp]
#include<stdio.h>
#include<math.h>
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<stdio.h>
#include<math.h>
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;
}
补充：软件开发 , C++ ,
