[湖大热身赛]Tractor（数学？)_the input starts with a single integer n (1 ≤ n ≤ -优快云博客

本文链接：https://blog.youkuaiyun.com/modiz/article/details/44261927

这篇博客探讨了Bessie如何在FJ的农场中移动并到达内部点的问题，通过分析其独特的移动规则来计算能够到达的点的数量。通过简化问题和找到规律，作者提供了一种解决方法来计算农场内可能到达的点数。

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Tractor

Time Limit: 6000ms, Special Time Limit:15000ms, Memory Limit:65536KB

Total submit users: 15, Accepted users: 13

Problem 13241 : No special judgement

Problem description

Bessie the Cow has stolen Farmer John’s tractor and is running wild on the coordinate plane! She, however, is a terrible driver, and can only move according to the following rules:

1. Each of her movements is in the same direction as either the positive x-axis or the positive y-axis.

2. Her nth movement takes her 2^(n−1) units forward in her chosen direction. (On her first movement, n = 1, so she moves 1 unit.)

Farmer John’s farm is on the coordinate plane, in the shape of a rectangle with corners at (0,0), (A,0), (0,B) and (A,B). If Bessie starts at (0,0), how many points inside the farm, including the boundary, could she reach?

Input

The input begins with an integer N (1 ≤ N ≤ 100) on a line by itself, indicating the number of test cases that follow. Each of the following N lines contains two space separated integers A and B (1 ≤ A,B ≤ 10^8), describing the upper-right corner of Farmer John’s farm.

Output

Output N lines, with the Nth line containing the number of points that Bessie could possibly reach in the Nth test case.

In the first test case of the sample, Bessie can reach the following six points: (0,0), (0,1), (1,0), (1,2), (2,1) and (3,0).

Sample Input

2
2 3
7 7

Sample Output

6
15

刚开始这题给的是错误的。说什么第n次移动是2*n-1，纠结题意了半天，后来darkdream发现其实是2^(n-1)好吧。然后就找规律了。

想着脑袋有点懵，但是感觉还是找对了方向，仔细推一下还是能解出来的，可能不是最优的那种把。

矩形取最短边，然后看最接近2^(n)-1，然后这里就是一个完整的二叉树=2*t+1;(t是边长了)，然后看最短边还剩下几个，肯定每条边上是能到一个点的，然后就是长边减去二叉树叶子节点的个数，然后以2的次方递增能加上几个叶子节点+最短边剩下的边的个数，好吧，解释的有点乱，估计也就我自己能看懂了。但是最后还是AC了。这种方法也没错。据说应该有更优的把。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <queue>
#include <map>
#include <stack>
#include <list>
#include <vector>
#include <ctime>
#define LL __int64
#define eps 1e-8
#define pi acos(-1)
using namespace std;
int main(){
	int T,a,b,i;
	scanf("%d",&T);
	while (T--){
		int ans=0;
		scanf("%d%d",&a,&b);
		if (a>b){
			a=a+b;
			b=a-b;
			a=a-b;
		}
		int g=a;
		int t=0;
		int s=1;
		for (i=1;g>=s;i++){
			g-=s;
			s*=2;
			t++;
		}
		ans=(s-1)*2+1+g;
		b=b-(s-1);
		int k=s;
		while (b&&k){
			ans++;
			b--;
			k--;
		}
		k=s+g;
		s*=2;
		while(b){
			if (b>=s){
				ans+=k;
				b-=s;
			}
			else {
				if (b+k>s) ans=ans+(b+k-s);
				break;
			}
			s*=2;
		}
		cout<<ans<<endl;
	}
	return 0;
}