计蒜客————Folding

•  30.64%
•  2000ms
•  262144K

As you can remember, Alex is fond of origami. She switched from squares to rectangles, and rectangles are much more difficult to master. Her main interest is to determine what is the minimum possible number of folds required to transform W ×H rectangle to w × h one. The result of each fold should also be 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.

Input

The first line of the input contains two integers Wand H — the initial rectangle dimensions. The second line contains two more integers w and h — the target rectangle dimensions (1 ≤ W, H, w, h ≤ 10^910​9​​).

Output

Output a single integer — the minimum required number of folds to transform the initial rectangle to the target one.

If the required transformation is not possible, output −1. 

样例输入1

2 72 2

样例输出1

2

样例输入2

10 64 8

样例输出2

2

样例输入3

5 51 6

样例输出3

-1

上面所给矩形折叠成下边所给矩形的最小折叠次数..模拟一下就好.

#include<iostream>#include<cstdio>#include<algorithm>using namespace std;int func2(int x1,int x2){    if(x2==x1) return 0;    for(int i=1;;i++){        x2*=2;        if(x2>=x1) return i;    }}int func1(int x1,int y1,int x2,int y2) {    if(x1<x2||y1<y2) return -1;    return func2(x1,x2)+func2(y1,y2);}int main() {    int x1,x2,y1,y2;    while(~scanf("%d%d%d%d",&x1,&y1,&x2,&y2)){        int t1=func1(x1,y1,x2,y2),t2=func1(x1,y1,y2,x2);        if(t1==-1||t2==-1)            printf("%d\n",max(t1,t2));        else printf("%d\n",min(t1,t2));    }    return 0;}