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  • Codeforces Round #295 (Div. 2) B. Two Buttons

    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.

    题意:

    给两个数,n,m,让n通过给点两种走步方法等于m

    对n可以有两种操作:

    1. 减1

    2.乘2

    我只会BFS。。。

     1 #include<iostream>
     2 #include<stdio.h>
     3 #include<math.h>
     4 #include<stdlib.h>
     5 #include<string.h>
     6 #include<queue>
     7 using namespace std;
     8 const int MAX=2e4+4;
     9 const int MIN=0;
    10 struct stu
    11 {
    12     int num;
    13     int step;
    14 }start,one,two;
    15 queue<struct stu>q;
    16 bool vis[MAX+5];
    17 void bfs(int n,int m)
    18 {
    19     memset(vis,false,sizeof(vis));
    20     start.num = n;
    21     start.step = 0;
    22     vis[n]=1;
    23     q.push(start);
    24     while(!q.empty())
    25     {
    26         one=q.front();
    27         two=q.front();
    28         q.pop();//删队首
    29         one.num*=2;
    30         one.step++;
    31         two.num-=1;
    32         two.step++;
    33         if(one.num==m)
    34         {
    35             printf("%d
    ",one.step);
    36             break;
    37         }
    38         if(two.num==m)
    39         {
    40             printf("%d
    ",two.step);
    41         }
    42         if(one.num>0 && one.num<MAX && !vis[one.num])
    43         {
    44             q.push(one);
    45             vis[one.num]=1;
    46         }
    47         if(two.num>0 && two.num<MAX && !vis[two.num])
    48         {
    49             q.push(two);
    50             vis[two.num]=1;
    51         }
    52     }
    53 }
    54 
    55 int main()
    56 {
    57     int n,m;
    58     while(~scanf("%d%d",&n,&m))
    59     {
    60         if(n>m)
    61         {
    62             printf("%d
    ",n-m);
    63         }
    64         else if(n==m)
    65         printf("0
    ");
    66         else
    67         {
    68             bfs(n,m);
    69         }
    70     }
    71     return 0;
    72 }
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  • 原文地址:https://www.cnblogs.com/xuesen1995/p/4332523.html
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