CF 520 B——Two Buttons————【広く探したり法則を探したりする】

J - Two Buttons
Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u
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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.
 
エラーポイント:タグ配列は付けられず、データ制限範囲は見られません.
 
広捜:

#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;

}

法則:逆思考ができる.n−1はある意味m+1に等しく、n*2はm/2に等しい.

#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;

}