快速切题sgu126. Boxes

126. Boxes

time limit per test: 0.25 sec. 
memory limit per test: 4096 KB

There are two boxes. There are A balls in the first box, and B balls in the second box (0 < A + B < 2147483648). It is possible to move balls from one box to another. From one box into another one should move as many balls as the other box already contains. You have to determine, whether it is possible to move all balls into one box.

Input

The first line contains two integers A and B, delimited by space.

Output

First line should contain the number N - the number of moves which are required to move all balls into one box, or -1 if it is impossible.

Sample Input

Sample Output

2 6

Sample Output

2

思路 1 只记录两个箱子的差值X,那么X=0的时候就能1次成功了
2 X的转移是X=SUM-2*X或者X=2*X-SUM(也即绝对值,X>0),注意到每次分母扩2倍,也就是说不超过32次
之后随便写一写就好啦,这个纯属胡整,我要睡觉!

#include <cstdio>
#include <cstring>
#include <cmath>
#include <queue>
#include <set>
using namespace std;
int a,b,sum;
queue<int >que;
set <int >st;
int main(){
    scanf("%d%d",&a,&b);
    if(a>b)swap(a,b);
    if(a==0||b==0)puts("0");
    else if(a==b)puts("1");
    else if((b-a)*2==a+b)puts("2");
    else if((b+a)&1)puts("-1");
    else {
        int sub=b-a;
        sum=a+b;
        st.insert(sub);
        que.push(sub);
        int step=0;
        while(!que.empty()){
            sub=que.front();que.pop();step++;
            if(sub==0||step>64)break;
            int t=abs(sum-2*sub);
            if(st.count(t)==0){que.push(t);st.insert(t);}
        }
        if(sub==0)printf("%d\n",step);
        else puts("-1");
    }
    return 0;
}