zoukankan      html  css  js  c++  java
  • UVA 12627 Erratic Expansion

    https://vjudge.net/problem/UVA-12627

    题目

    Piotr found a magical box in heaven. Its magic power is that if you place any red balloon inside it then, after one hour, it will multiply to form 3 red and 1 blue colored balloons. Then in the next hour, each of the red balloons will multiply in the same fashion, but the blue one will multiply to form 4 blue balloons. This trend will continue indefinitely.
    The arrangements of the balloons after the 0-th, 1-st, 2-nd and 3-rd hour are depicted in the following diagram.

    As you can see, a red balloon in the cell (i, j) (that is i-th row and j-th column) will multiply to produce 3 red balloons in the cells $(i ast 2 - 1, j ast 2 - 1)$, $(i ast 2 - 1, j ast 2)$, $(i ast 2, j ast 2 - 1)$ and a blue balloon in the cell $(i ast 2, j ast 2)$. Whereas, a blue balloon in the cell $(i, j)$ will multiply to produce 4 blue balloons in the cells $(i ast 2 - 1, j ast 2 - 1)$, $(i ast 2 - 1, j ast 2)$, $(i ast 2, j ast 2 - 1)$ and $(i ast 2, j ast 2)$. The grid size doubles (in both the direction) after every hour in order to accommodate the extra balloons.

    In this problem, Piotr is only interested in the count of the red balloons; more specifically, he would like to know the total number of red balloons in all the rows from A to B after K-th hour.

    Input

    The first line of input is an integer $T$ $(T < 1000)$ that indicates the number of test cases. Each case contains 3 integers $K$, $A$ and $B$. The meanings of these variables are mentioned above. $K$ will be in the range $[0, 30]$ and $1 leq A leq B leq 2^K$ .

    Output

    For each case, output the case number followed by the total number of red balloons in rows $[A,B]$ after $K$-th hour.

    Sample Input

    3
    0 1 1
    3 1 8
    3 3 7
    

     Sample Output

    Case 1: 1
    Case 2: 27
    Case 3: 14
    

     题解

    心情不太好,不解释了

    AC代码

    #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
    #include<bits/stdc++.h>
    using namespace std;
    #define REP(r,x,y) for(register int r=(x); r<(y); r++)
    #define REPE(r,x,y) for(register int r=(x); r<=(y); r++)
    #ifdef sahdsg
    #define DBG(...) printf(__VA_ARGS__)
    #else
    #define DBG(...)
    #endif
    
    #define MAXN 10000007
    typedef long long LL;
    
    template <class T>
    inline void read(T& x) {
    	char c=getchar();
    	int f=1;
    	x=0;
    	while(!isdigit(c)&&c!='-')c=getchar();
    	if(c=='-')f=-1,c=getchar();
    	while(isdigit(c)) {
    		x=x*10+c-'0';
    		c=getchar();
    	}
    	x*=f;
    }
    template <class T>
    void write(T x) {
    	if(x<0) {
    		putchar('-');
    		write(-x);
    		return;
    	}
    	if(x>9) write(x/10);
    	putchar(x%10+'0');
    }
    
    long long f1[31]={1,3};
    inline void getf1() {
    	REPE(i,2,30) {
    		f1[i]=f1[i-1]*3;
    	}
    }
    inline long long c(int i) {
    	return 1<<i;
    }
    
    inline long long f(int k, int i) {
    	if(k==0) {
    		if(i<=1) return 1;
    		else return 0;
    	}
    	if(i<=c(k-1))
    		return f(k-1,i)*2+f1[k-1];
    	else if(i<=c(k))
    		return f(k-1,i-c(k-1));
    	else return 0;
    }
    int main() {
    	#ifdef sahdsg
    	freopen("in.txt", "r", stdin);
    	#endif
    	getf1();
    	int n;
    	read(n);
    	
    	REPE(i,1,n) {
    		int k,a,b;
    		read(k);read(a);read(b);
    		printf("Case %d: ", i);
    		write(f(k,a) - f(k,b+1)); putchar('
    ');
    	}
    	return 0;
    }
    
  • 相关阅读:
    Android之ListView中的分割线
    Android 数据存储(XML解析)
    Altium Designer 15 --- Nets Update
    IAR ------ 基本使用
    Altium Designer 15 --- Make LOGO/ICON
    点云数据处理
    Reconstruction(三维重建)文件被修改
    ORB_SLAM2应用实践_ROS小强机器人
    OpenCV4Android安装
    开题报告
  • 原文地址:https://www.cnblogs.com/sahdsg/p/10490262.html
Copyright © 2011-2022 走看看