GDUT_排位赛题解报告_第1场_B. MooBuzz

题目：
Farmer John’s cows have recently become fans of playing a simple number game called “FizzBuzz”. The rules of the game are simple: standing in a circle, the cows sequentially count upward from one, each cow saying a single number when it is her turn. If a cow ever reaches a multiple of 3, however, she should say “Fizz” instead of that number. If a cow reaches a multiple of 5, she should say “Buzz” instead of that number. If a cow reaches a multiple of 15, she should say “FizzBuzz” instead of that number. A transcript of the first part of a game is therefore: 1, 2, Fizz, 4, Buzz, Fizz, 7, 8, Fizz, Buzz, 11, Fizz, 13, 14, FizzBuzz, 16

Having a slightly more limited vocabulary, the version of FizzBuzz played by the cows involves saying “Moo” instead of Fizz, Buzz, and FizzBuzz. The beginning of the cow version of the game is therefore

1, 2, Moo, 4, Moo, Moo, 7, 8, Moo, Moo, 11, Moo, 13, 14, Moo, 16

Given $$N$$($1\le N\le 10^9$), please determine the $$N$$th number spoken in this game.

Test cases 2-5 satisfy $$N\le 10^6$$.

Input
The input consists of a single integer, $$N$$.

Output
Please print out the Nth number spoken during the game.

Example
inputCopy
4
outputCopy
7
Note
The 4th number spoken is 7. The first 4 numbers spoken are 1, 2, 4, 7, since we skip over any time a cow says “Moo”.

这活指的是逢3 5倍数的数都会抹去，问你第N个幸存者是谁。 这个题目做法有很多，比较简单我认为应该是二分还有预处理了，之前遇到这样的题，那时候不会做二分，于是预处理了一下，但是现在数据加强，直接上二分，就可以

对幸存数字进行二分：

	LL left=0,right=1e18;
	while(left+1<right)
	{
		LL mid=(left+right)/2;
		if(mid-mid/3-mid/5+mid/15==n)
		{
			while(mid%3==0||mid%5==0)mid--;
			printf("%lld\n",mid);
			break;
		}
		else if(mid-mid/3-mid/5+mid/15<n)left=mid;
		else right=mid;
	}

代码比较少，但是自己对二分这样的东西有点不太自然，微调的时候总是要调半天，希望做多一点题目，然后回顾博客的二分题目来加深亲切感。