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 two. After clicking the blue button, device subtracts one from the number 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 has to make in order to achieve this result?
The first and the only line of the input contains two distinct integers n and m (1 ≤ n, m ≤ 104), separated by a space .
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.
4 6
2
10 1
9
In 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 is unnecessary, so we need to push the blue button nine times.
本题的题意，有台带两个不同颜色按钮的装置，按红色按钮对当前数乘2，按蓝色按钮对当前数减1，且得到的数为正数。
给定一个数n，用该装置进行操作，进行若干次操作后使该数正好等于数m，问所需最少的操作步数。
这题我的想法是用bfs来做，网上有说用贪心做，没细看。前几次交的代码各种tlm，没进行良好的剪枝。
#include<iostream> #include<cstring> #include<queue> #include<cstdio> using namespace std; const int MAX_N = 20010; int vis[MAX_N]; int main(){ // freopen("D:\\in.txt","r",stdin); // freopen("D:\\out.txt","w",stdout); memset(vis,0,sizeof(vis)); queue<int> q; int n,m; cin>>n>>m; bool ans = false; vis[n] = 1; q.push(n); if(n >= m) cout<<n-m<<endl; else{ while(q.size()){ int a = q.front(); q.pop(); if(a == m){ ans = true; break; } int temp = a*2; if(temp > 0 && temp < m * 2 && vis[temp]==0){ vis[temp] = vis[a]+1; q.push(temp); } temp = a-1; if(temp > 0 && temp < m * 2 && vis[temp]==0){ vis[temp] = vis[a]+1; q.push(temp); } } } if(ans) cout<<vis[m]-1<<endl; return 0; }