Codeforces 520B Buttons

Time limit:2000MS Memory Limit:262144KB 64bit IO Format:%i64d & %i64u

Description

Vasya has found a strange device. On the front panel of a device there are:a red button, a blue button and a display showing some positive integer. After clicking the red button, device multiplies the displayed number by. After clicking the blue button, device subtracts one is on the display. If at some point the number stops being positive, the device breaks down. The display can show arbitrarily large numbers. Initially, the display shows number n.

Bob wants to get number m on the display. What minimum number of clicks he have to do in order to achieve this result?

Input

The first and the only line of the input contains, distinct integers n and m (1≤ n , m ≤10^{4), separated by a space.}

Output

Print a single number-the minimum number of times one needs to push the button required to get the number m out of number n.

Sample Input

Input

4 6

Output

2

Input

10 1

Output

9

Hint

The first example you need to push the blue button once, and then push the red button once.

In the second example, doubling the number was unnecessary, so we need to push the blue button nine times.

Train of thought 1:math for n < m, starting from M, if M is even, M is halved, otherwise, plus 1 is halved (corresponding result plus 1) until M <= N

#include <cstdio>#include<cstring>#include<iostream>using namespacestd;intN, M;voidsolve () {cin>> N >>m; if(n >= m) {cout << n-m << Endl;return;} intAns =0;  while(N <m) {if(M &1) {ans++;m++;} M>>=1; Ans++; } ans+ = n-m; cout<< ans <<Endl;}intMain () {solve (); return 0;}

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Thinking 2:bfs + pruning (memory)

/*Times memy 78ms 2104k by orc*/#include<iostream>#include<algorithm>#include<queue>#include<cstring>using namespacestd;intN, M;BOOLvis[20005];//the VIS tag array here is not the same as the previous mark has been done and do not go repeat pathstructnode{//Here's a memo, a memory search .    intx; intcnt;};voidBFS (node v) {memset (Vis,false,sizeofvis); Queue<node>que;    Que.push (v);    node now, NEX;  while(!Que.empty ()) { Now=Que.front ();        Que.pop (); if(vis[now.x])Continue; if(Now.x <=0)Continue;//since we have to make a memo, the subscript can't be 0 .        if(Now.x > m) {//important pruning, if now.x > m, that is now.x * 2, there is no need to queue, only through-to reach the status mnex.x = now.x-1; Nex.cnt= now.cnt +1;                Que.push (NEX); Continue; }        if(now.x = = m) {cout << now.cnt << Endl;return;} nex.x= now.x-1; Nex.cnt= now.cnt +1;        Que.push (NEX); nex.x= now.x *2; Nex.cnt= now.cnt +1;        Que.push (NEX); Vis[now.x]=true;//Loop at the end of the current outbound element now tag    }}intMain () {CIN>> N >>m; if(n >= m) cout << n-m <<Endl; Else{node V; v.x=N; V.cnt=0;    BFS (v); }}

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Codeforces 520B Buttons