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2.8 kB
| // Stockholm | |
| // Solution by Jacob Plachta | |
| using namespace std; | |
| template<typename T> T Abs(T x) { return(x<0 ? -x : x); } | |
| template<typename T> T Sqr(T x) { return(x*x); } | |
| string plural(string s) { return(Sz(s) && s[Sz(s)-1]=='x' ? s+"en" : s+"s"); } | |
| const int INF = (int)1e9; | |
| const LD EPS = 1e-12; | |
| const LD PI = acos(-1.0); | |
| bool Read(int &x) | |
| { | |
| char c,r=0,n=0; | |
| x=0; | |
| for(;;) | |
| { | |
| c=GETCHAR(); | |
| if ((c<0) && (!r)) | |
| return(0); | |
| if ((c=='-') && (!r)) | |
| n=1; | |
| else | |
| if ((c>='0') && (c<='9')) | |
| x=x*10+c-'0',r=1; | |
| else | |
| if (r) | |
| break; | |
| } | |
| if (n) | |
| x=-x; | |
| return(1); | |
| } | |
| pair<int,bool> Cost(string s) | |
| { | |
| if (!Sz(s)) | |
| return(mp(0,0)); | |
| int i=Sz(s)-1; | |
| // delete equal digits from the end | |
| while ((i) && (s[i]==s[i-1])) | |
| i--; | |
| // delete one more digit from the end | |
| if (i) | |
| i--; | |
| // now empty? | |
| if (!i) | |
| return(mp(0,Sz(s)>=2 && s[0]==s[Sz(s)-1])); | |
| pair<int,bool> c=Cost(s.substr(0,i)); | |
| c.x++; | |
| return(c); | |
| } | |
| int main() | |
| { | |
| if (DEBUG) | |
| freopen("in.txt","r",stdin); | |
| // vars | |
| int T,t; | |
| LL A,B; | |
| string X,Y; | |
| int p,ans; | |
| pair<int,bool> c1,c2; | |
| // testcase loop | |
| Read(T); | |
| Fox1(t,T) | |
| { | |
| // input | |
| scanf("%lld%lld",&A,&B); | |
| // convert nodes to binary | |
| A++,B++; | |
| X=Y=""; | |
| while (A) | |
| X=char(A%2+'0')+X,A/=2; | |
| while (B) | |
| Y=char(B%2+'0')+Y,B/=2; | |
| // find LCA | |
| Fox(p,min(Sz(X),Sz(Y))) | |
| if (X[p]!=Y[p]) | |
| break; | |
| // compute costs to LCA | |
| X=X.substr(p); | |
| Y=Y.substr(p); | |
| c1=Cost(X),c2=Cost(Y); | |
| ans=c1.x+c2.x; | |
| // at least one node is coming from the outside, and neither is the LCA? | |
| if ((Sz(X)) && (Sz(Y)) && (c1.y || c2.y)) | |
| ans++; | |
| // output | |
| printf("Case #%d: %d\n",t,ans); | |
| } | |
| return(0); | |
| } |