CF520B——Two Buttons——————【广搜或找规律】

J - Two Buttons

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 520B

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 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.

Sample Input

Input

4 6

Output

2

Input

10 1

Output

9

Hint

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.

出错点：没加标记数组，没有看数据限制范围。

广搜：

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;
struct node{

    int out;
    int dep;
}pos;
const int maxn=11000;
bool mark[maxn];
queue<node>Q;
int aim;
int bfs(int n){

    memset(mark,0,sizeof(mark));
    pos.out=n;
    pos.dep=0;
    Q.push(pos);
    mark[pos.out]=1;
    while(!Q.empty()){

        node tmp=Q.front();
        Q.pop();
        if(tmp.out!=aim){

           node tmp1,tmp2;
           tmp1.dep=tmp.dep+1;
           tmp1.out=tmp.out*2;
           if(tmp1.out>=1&&tmp1.out<=10000&&!mark[tmp1.out]){
              
                Q.push(tmp1);
                mark[tmp1.out]=1;
           }
            tmp2.out=tmp.out-1;
            tmp2.dep=tmp.dep+1;
            if(tmp2.out>=1&&tmp2.out<=10000&&!mark[tmp2.out]){
                
                Q.push(tmp2);
                mark[tmp2.out]=1;
            }
        }else{

            return tmp.dep;
        }
    }
}
int main(){

    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF){

        while(!Q.empty()){

            Q.pop();
        }

        aim=m;
        int ans;
        if(n==aim){

            printf("0
");
        }else{

            ans=bfs(n);
            printf("%d
",ans);
        }
    }
    return 0;
}

规律：可以逆向思考。n-1在某种意义上等同于m+1，n*2等同于m/2。

#include<stdio.h>
int main(){

    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF){

        int cnt=0;
        if(n>=m){

            printf("%d
",n-m);
        }else{

            while(n!=m){

                if(m&1){

                    m+=1;
                    cnt++;
                }
                m/=2;
                cnt++;
                if(n>m){

                    cnt+=n-m;
                    break;
                }
            }
            printf("%d
",cnt);
        }
    }
    return 0;
}