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Circle City
Circle City
+ 0 comments Passed all but one test case:
// Complete the solve function below. static String solve(int d, int k) { //get integer format of radius int r = (int)Math.sqrt(d); int count = 0; //loop for all integer values on X axis for(int i = -r; i<r; i++){ //calculate the float/double value of Y // Let's say, it could be 2.0 or 2.3 double y = Math.sqrt(d - i*i); // convert to INT so it will become either 2.0 in above cases int intY = (int)y; //now compare the int version with original version. //if they are same, then Y is also integer and hence we got our point on //the circle. Count is incremented by 2 because, this new point is applicable both above and below x axis. if((float)intY == y ){ count+=2; } } if(k >= count){ return "possible"; } else { return "impossible"; } }
+ 0 comments some tese case fail this code .....can you tell why that.....??
int solve(long int d,long int k) { int r=sqrt(d); int t,p,ans=4; //char *possible,*impossible;
if((r==1 && k>=4) || (r==2 && k>=4)) return 1; if(r>2) { p=2,t=2; while((r-2)>2 && (r-t)>=2) { p=floor((r-t)/2); t=(t+p); ans+=4; } } if(ans<=k) return 1; else return 0;
}
+ 0 comments //THis code clear all test cases
static String solve(int d, int k)
{ long count=0; for(long i=0;i<Math.sqrt(d);i++) { long p=(long)Math.sqrt((long)d-i*i); if((p*p+i*i)==d) { count+=4; } } if(k>=count) { return "possible"; } return "impossible"; }
+ 0 comments can anyone help me out... i get TLE on testcase 2. Is my code efficient?
import math t=int(input()) for _ in range(t): d,k=[int(x) for x in input().split()] ans=0 r=math.ceil(d**.5) for i in range(0,r): x=(d-i**2)**.5 #y=int(math.sqrt(d-i**2)) if x==int(x): ans+=4 #print(i,y) if k<ans: print("impossible") else: print("possible")
+ 0 comments Could be solved easily with https://en.wikipedia.org/wiki/Fermat%27s_theorem_on_sums_of_two_squares and where = number of decompositions of in two squares. The French version of the above page gives this formula and the meaning of and .
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