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#include <iostream> #include <cstdio> #include <cmath> #include <vector> #define ll long long #define MOD 1000000007 using namespace std; int _pow(int n,int p) { if(p==0) return 1; else return n*_pow(n,p-1); } int isAnyRoot(int a,int b) { if(a%b!=0) return -1; int cnt=0; while(a>=b) { cnt+=1; a=a/b; if(a>=b && a%b!=0) return -1; } return cnt; } int main() { //cout<<_pow(6,2)<<"\n"; int H,W,m,n,non_working_cat; int c1,c2; int total_non_working_cat , height_of_the_cat_stack , height_of_each_internal_cat; while(scanf("%d%d",&H,&W),H!=0,W!=0) { if(H==1) { printf("%d %d\n",0,H); continue; } //that means there is one working cat ; so there can be two possible scenarios //1. if the height of the initial cat is nth power of 2, //i.e. just next(inside of which hat working cat is staying) non working cat //whose height should be 2 so that 2/N+1==1; // In this acse total non working cat will be n (2^n=initial height), and height of the cat stack = 2+4+...2^n //2. if the height of the initial cat is not nth power of 2; then 1=(Height)/N+1, so N+1=Height, //means total cat= Height of the initial cat //only one is non working rest are working cat //and total height should be (N*1+initial height) or (N*1+1) if they are considered in a same ground if(W==1) { m = isAnyRoot(H,2); if(m!=-1) { printf("%d %d\n",m,_pow(2,m+1)-1); } else { printf("%d %d\n",1,2*H-1); } continue; } /* m=0; while(++m) { n = pow(W*1.0,1.0/m); if(H<=pow(n+1, m)) break; }*/ n=1; while(1) { n+=1; c1 = isAnyRoot(W,n); c2 = isAnyRoot(H,n+1); if(c1==c2 && c1!=-1) break; } m=c1; total_non_working_cat = 0; height_of_the_cat_stack = 0; for(int i=0;i<m;i++) { non_working_cat = _pow(n,i); total_non_working_cat += non_working_cat; height_of_each_internal_cat = (H/_pow(n+1,i)); height_of_the_cat_stack += (non_working_cat*height_of_each_internal_cat); } height_of_the_cat_stack += W;//as working cats are of height 1. printf("%d %d\n",total_non_working_cat,height_of_the_cat_stack); } return 0; } |