mes pie The san oat position out float transform

B. Two Buttons time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

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?

Input

The first and the only line of the input contains two distinct integers n and m (1?≤?n,?m?≤?104), 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.

Examples input Copy

4 6

output Copy

2

input Copy

10 1

output Copy

9

Note

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.

本題的題意，有臺帶兩個不同顏色按鈕的裝置，按紅色按鈕對當前數乘2，按藍色按鈕對當前數減1，且得到的數為正數。

給定一個數n，用該裝置進行操作，進行若幹次操作後使該數正好等於數m，問所需最少的操作步數。

這題我的想法是用bfs來做，網上有說用貪心做，沒細看。前幾次交的代碼各種tlm，沒進行良好的剪枝。

#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;
}

Codeforces Round #295 (Div. 2)