2431: Shift and Increment

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探讨了如何通过移位和增1操作实现从任意数到0的转换，并寻找最短的操作路径。此问题涉及CPU中算术逻辑单元的基本操作。

2431: Shift and Increment

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	5s	16384K	1132	174	Standard

Shift and increment is the basic operations of the ALU (Arithmetic Logical Unit) in CPU. One number can be transform to any other number by these operations. Your task is to find the shortest way from x to 0 using the shift (*=2) and increment (+=1) operations. All operations are restricted in 0..n, that is if the result x is greater than n, it should be replace as x%n.

Input and Output

There are two integer x, n (n <=1000000)

Sample Input

2 4
3 9

Sample Output

1
3

Problem Source: provided by skywind

#include<iostream>
#include<algorithm>
using namespace std;
const int inf=1000005;
bool flag[inf];
int a[inf];
int main()
{
    int x,n,i;
    while(scanf("%d%d",&x,&n)==2)
    {
        memset(flag,0,sizeof(flag));
        int begin,end,num;
        begin=end=num=0;
        a[num++]=x;
        flag[x]=1;
        int count=0;
        while(flag[0]==0)
        {
            end=num;
            count++;
            for(i=begin;i<end;i++)
            {
                int t=(a[i]*2)%n;
                if(flag[t]==0)
                {
                    flag[t]=1;
                    a[num++]=t;
                }
                t=(a[i]+1)%n;
                if(flag[t]==0)
                {
                    flag[t]=1;
                    a[num++]=t;
                }
            }
            begin=end;
        }
        printf("%d/n",count);
    }
    return 0;
}