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  • AcWing98 分形之城 (递归)

    题目链接:https://www.acwing.com/problem/content/100/

    为方便运算,城市编号从(0)开始,
    按城市分级递归求解(D)编号城市的坐标
    注意变换后坐标的计算(坐标从((0,0))开始)

    四舍五入用(%0lf)即可

    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<iostream>
    #include<cmath>
    #include<stack>
    #include<queue>
    using namespace std;
    typedef long long ll;
    typedef unsigned long long ull;
    typedef pair<ll,ll> P;
    
    const int maxn = 40;
    
    int T;
    int N;
    ll A,B;
    
    double dis(P a, P b){ return (double)sqrt(1.0 * (a.first - b.first) * (a.first - b.first) + 1.0 * (a.second - b.second) * (a.second - b.second)) * 10; }
    
    P dfs(int n,ll pos){
    	if(n == 0){ return make_pair(0, 0); }
    	ll len = 1ll << (n - 1), num = 1ll << (2 * n - 2);
    	P tp = dfs(n - 1, pos % num);
    	ll x = tp.first, y = tp.second;
    	
    	int d = pos / num;
    	if(d == 0) return make_pair(y, x);
    	if(d == 1) return make_pair(x, y + len);
    	if(d == 2) return make_pair(x + len, y + len);
    	if(d == 3) return make_pair(2 * len - y - 1, len - x - 1);
    }
    
    ll read(){ ll s=0,f=1; char ch=getchar(); while(ch<'0' || ch>'9'){ if(ch=='-') f=-1; ch=getchar(); } while(ch>='0' && ch<='9'){ s=s*10+ch-'0'; ch=getchar(); } return s*f; }
    
    int main(){
    	T = read();
    	while(T--){
    		N = read(), A = read(), B = read(); --A, --B;
    		++N;
    		
    		ll num = 1ll << (2 * N - 2);
    		P a = dfs(N - 1, A % num);
    		P b = dfs(N - 1, B % num);
    //		printf("%lld %lld %lld %lld
    ",a.first, a.second, b.first, b.second); 
    		printf("%.0lf
    ",dis(a, b));
    	}
    	
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/tuchen/p/13909922.html
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