Gym - 101142F Folding（折半）

传送门

As you can remember, Alex is fond of origami. She switched from squares to rectangles, and rectanglesare much more difficult to master. Her main interest is to determine what is the minimum possiblenumber of folds required to transform W × H rectangle to w × h one. The result of each fold should alsobe rectangular, so it is only allowed to make folds that are parallel to the sides of the rectangle.Help Alex and write a program that determines the minimum required number of folds.InputThe first line of the input contains two integers W and H — the initial rectangle dimensions. The secondline contains two more integers w and h — the target rectangle dimensions (1 ≤ W, H, w, h ≤ 10^9).OutputOutput a single integer — the minimum required number of folds to transform the initial rectangle tothe target one.If the required transformation is not possible, output −1.Examplesfolding.in 2 72 210 64 85 51 6folding.out22-1In the first example you should fold 2 × 7 rectangle to 2 × 4, and then to 2 × 2.In the second example you should fold 10 × 6 rectangle to 10 × 4, then to 8 × 4, and rotate it to 4 × 8.In the third example it is impossible to fold 5 × 5 rectangle to 1 × 6 one (remember that folds must beparallel to the rectangle sides).

题目大意：

给了一个 W∗H 的大矩形和一个 w∗h 的小矩形，现在让你将这个大矩形进行折叠使其变成这个小矩形，求需要的最小步数（要求是这个小矩形不能在进行拆分了），如果不能完成转化输出 -1.

样例解释：
Input：

2 7
2 2

Output：
2

首先将 2∗7 的矩形折叠成 2∗4的矩形，然后在将 2∗4 的矩形折叠成 2∗2 的，不能上来就将 2∗7 的矩形折叠成 2∗2的因为还有 2∗5的这是不允许的。

解题思路：

首先我们很容易想到是 −1 的情况：就是当 max(w,h)>max(W,H) 的时候就输出 −1；然后在来考虑能够折叠的情况，其实还是很容易想到的是折半来分，这样是最小的步数，然后需要满足的条件就是不能有比这个大的，那么如果当前边是奇数的话就 +1 然后除以 2，否则直接除以 2,然后我们就将(W,w)和(H,h)都分别算一遍，然后在将 (H,w)和 (W,h)算一遍取最小值就 OK了。
My Code：

#include <iostream>#include <cstdio>#include <cstdlib>#include <cstring>using namespace std;typedef long long LL;const LL INF = 1e18+5;LL Solve(LL mx, LL mn){    if(mn > mx)        return INF;    LL cnt = 0;    while(mx > mn){        mx = (mx+1)>>1LL;        cnt++;    }    return cnt;}int main(){    ///freopen("folding.in","r",stdin);    ///freopen("folding.out","w",stdout);    LL w, h ,ww, hh;    while(cin>>w>>h>>ww>>hh){        LL TW = max(w,h);        LL TH = min(w,h);        LL tw = max(ww,hh);        LL th = min(ww,hh);        if(tw > TW){            puts("-1");            continue;        }        if(th > TH){            puts("-1");            continue;        }        ///cout<<max(Solve(TW, tw)+Solve(TH,th),Solve(TW,th)+Solve(TH,tw))<<endl;        cout<<min(Solve(TW, tw)+Solve(TH,th),Solve(TW,th)+Solve(TH,tw))<<endl;    }    return 0;}