2016-2017 ACM-ICPC, NEERC, Northern Subregional Contest F（gym/101142 F）

gym/101142 F

题意：给一张长宽W*H的纸， 求最少多少次可以折成长宽 w*h，每次只能平行边缘折

思路：每次对折是最优的，最后一次不一定对折，但是只计算次数，可以不用管最后一次到底怎么折的，只要第一次折到W<=w H<=h，即可，一共有2种情况：1、W->w , H->h，2、W->h, H->w； 2种情况都要考虑

AC代码：

#include "iostream"
#include "iomanip"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#pragma comment(linker, "/STACK:102400000,102400000")
#define bug(x) cout<<x<<" "<<"UUUUU"<<endl;
#define mem(a,x) memset(a,x,sizeof(a))
#define step(x) fixed<< setprecision(x)<<
#define mp(x,y) make_pair(x,y)
#define pb(x) push_back(x)
#define ll long long
#define endl ("\n")
#define ft first
#define sd second
#define lrt (rt<<1)
#define rrt (rt<<1|1)
using namespace std;
const ll mod=1e9+7;
const ll INF = 1e18+1LL;
const int inf = 1e9+1e8;
const double PI=acos(-1.0);
const double eps=1e-9;
const int N=1e5+100;

int main(){
    freopen("folding.in","r",stdin); 
　　 freopen("folding.out","w",stdout);
    double W,H,w,h;
    int ans1=0,ans2=0;
    cin>>W>>H>>w>>h;
    double W1=W,H1=H;
    if(W<H) swap(W,H);
    if(w<h) swap(w,h);
    if(W<w || H<h){
        cout<<"-1\n";
        return 0;
    }
    while(W-w>eps){
        W/=2;
        ans1++;
    }
    while(H-h>eps){
        H/=2;
        ans1++;
    }
    if(W1<H1) swap(W1,H1);
    if(w>h) swap(w,h);
    if(W1<w || H1<h) ans2=inf;
    while(W1-w>eps){
        W1/=2;
        ans2++;
    }
    while(H1-h>eps){
        H1/=2;
        ans2++;
    }
    int ans=min(ans1,ans2);
    cout<<ans<<endl;
    return 0;
}