Given coordinates of 2 points on a cartesian plane, output the distance between them rounded up to nearest integer.
Input:
The first line of the input contains the number of test cases T. Each line contains 4 integers denoting those 2 points as (x1,y1), (x2,y2).
Output:
For each test case print in a single line the distance between the two points.
Constraints:
1<=T<=100
-1000000<=|x1,x2,y1,y2|<=1000000
The first line of the input contains the number of test cases T. Each line contains 4 integers denoting those 2 points as (x1,y1), (x2,y2).
Output:
For each test case print in a single line the distance between the two points.
Constraints:
1<=T<=100
-1000000<=|x1,x2,y1,y2|<=1000000
Example:
Input:
4
0 0 2 -2
-20 23 -15 68
30 37 79 -51
-69 63 57 11
Output:
3
45
101
136
Input:
4
0 0 2 -2
-20 23 -15 68
30 37 79 -51
-69 63 57 11
Output:
3
45
101
136
C-Solution
#include <stdio.h>
#include <math.h>
int main() {
unsigned int t,x1,x2,y1,y2;float ans;
scanf("%d",&t);
while(t--)
{
scanf("%d",&x1);
scanf("%d",&x2);
scanf("%d",&y1);
scanf("%d",&y2);
x1=y1-x1;y1=y2-x2;
x2=x1*x1;y2=y1*y1;
// printf("\n%d %d %d %d\n",x1,x2,y1,y2);
ans=x2+y2;ans=(sqrt(ans));
printf("%.0f\n",ans);ans=0;
}
return 0;
}
#include <math.h>
int main() {
unsigned int t,x1,x2,y1,y2;float ans;
scanf("%d",&t);
while(t--)
{
scanf("%d",&x1);
scanf("%d",&x2);
scanf("%d",&y1);
scanf("%d",&y2);
x1=y1-x1;y1=y2-x2;
x2=x1*x1;y2=y1*y1;
// printf("\n%d %d %d %d\n",x1,x2,y1,y2);
ans=x2+y2;ans=(sqrt(ans));
printf("%.0f\n",ans);ans=0;
}
return 0;
}
No comments:
Post a Comment