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嵊州普及Day5T2

作者:互联网

题意:将(w,h)的纸条折成(W,H),最少需几步。

思路:横竖互不干扰,然后最多可折int型一半,拿个函数判断两次比较即可,然后折不了的条件是需要的矩形大于给的矩形。

见代码:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int w,h,W,H,ans=0,sum;
void fold(int h,int l,int x,int y)
{
    if(h>=x&&l>=y)
    {   
        if(h!=x)
        while(h-x>=h/2)
        {
            h-=h/2;ans++;
            if(h==x)
            break;
        }
        if(h!=x)
        ans++;
        if(l!=y)
        while(l-y>=l/2)
        {
            l-=l/2;ans++;
            if(l==y)
            break;
        }
        if(l!=y)
        {
         ans++;
        }
    }
}
int main()
{
    freopen("folding.in","r",stdin);
    freopen("folding.out","w",stdout);
    cin>>w>>h>>W>>H;
    int a=w,b=h;
    fold(w,h,W,H);
    w=a;
    h=b;
    sum=ans;
    ans=0;
    fold(w,h,H,W);
    if(sum!=0&&ans!=0)
    ans=min(sum,ans);
    else
    {
        if(ans==0)
        ans=sum;
    }
    if((w==W&&h==H)||(w==H&&h==W))
    {
        cout<<0<<endl;
        return 0;
    }
    if(ans==0)
    cout<<-1;
    else
    cout<<ans<<endl;
    return 0;
}

好题哉!!!

标签:嵊州,int,sum,普及,++,&&,ans,include,Day5T2
来源: https://www.cnblogs.com/qing1/p/11191373.html