HDU6768 The Oculus（Hash）

Let's define the Fibonacci sequence $F_1,F_2,\dots$ as $F_1=1,F_2=2,F_i=F_{i-1}+F_{i-2}$ ($i\geq 3$).

It's well known that every positive integer $x$ has its unique Fibonacci representation $(b_1,b_2,\dots,b_n)$ such that:

· $b_1\times F_1+b_2\times F_2+\dots+b_n\times F_n=x$.

· $b_n=1$, and for each $i$ ($1\leq i<n$), $b_i\in\{0,1\}$ always holds.

· For each $i$ ($1\leq i<n$), $b_i\times b_{i+1}=0$ always holds.

For example, $4=(1,0,1)$, $5=(0,0,0,1)$, and $20=(0,1,0,1,0,1)$ because $20=F_2+F_4+F_6=2+5+13$.

There are two positive integers $A$ and $B$ written in Fibonacci representation, Skywalkert calculated the product of $A$ and $B$ and written the result $C$ in Fibonacci representation. Assume the Fibonacci representation of $C$ is $(b_1,b_2,\dots,b_n)$, Little Q then selected a bit $k$ ($1\leq k<n$) such that $b_k=1$ and modified $b_k$ to $0$.

It is so slow for Skywalkert to calculate the correct result again using Fast Fourier Transform and tedious reduction. Please help Skywalkert to find which bit $k$ was modified.

Input

The first line of the input contains a single integer $T$ ($1 \leq T \leq 10\,000$), the number of test cases.

For each case, the first line of the input contains the Fibonacci representation of $A$, the second line contains the Fibonacci representation of $B$, and the third line contains the Fibonacci representation of modified $C$.

Each line starts with an integer $n$, denoting the length of the Fibonacci representation, followed by $n$ integers $b_1,b_2,\dots,b_n$, denoting the value of each bit.

It is guaranteed that:

· $1\leq |A|,|B|\leq 1\,000\,000$.

· $2\leq |C|\leq |A|+|B|+1$.

·$\sum |A|,\sum |B|\leq 5\,000\,000$.

Output

For each test case, output a single line containing an integer, the value of $k$.

Sample Input

1
3 1 0 1
4 0 0 0 1
6 0 1 0 0 0 1

Sample Output

4

学习一波QAQ

整体的思路大概就是设modify的是第k位，那么有：$A\times B=C+F_k$，移项后可得：$F_k=A\times B -C$，预处理出斐波那契数列后直接查找即可。但是注意到A,B都在1e6级别，这个范围有点大，因此需要找一个模数P，使得Fib[1]~Fib[2e6]这个范围内的元素模P两两不同余（为啥是Fib[2e6]呢？因为k的范围最大到2e6）。题解很巧妙地用了自然溢出，即P取unsigned long long的最大值$2^{64}$。

（题目里说不会有连续的1，又注意到这个数列是斐波那契数列，因此这个条件的含义就是一个数不会有其他的表示方法，比如（1110）和（1001）的值是一样的，而前者有连续的1，就不会出现在输入中。但这...会产生影响吗..）

正常的Hash参照大佬博客：https://www.cnblogs.com/stelayuri/p/13367956.html 可以用进阶map也可以用双Hash避免冲突。

#include <bits/stdc++.h>
using namespace std;
unsigned long long fib[2000005], A, B, C, k;
int main()
{
	int t;
	cin >> t;
	fib[1] = 1, fib[2] = 2;
	for(int i = 3; i <= 2000000; i++) fib[i] = fib[i - 1] + fib[i - 2];
	while(t--)
	{
		int na, nb, nc;
		A = B = C = 0;
		cin >> na;
		for(int i = 1; i <= na; i++)
		{
			int temp = 0;
			scanf("%d", &temp);
			if(temp) A += fib[i];
		}
		cin >> nb;
		for(int i = 1; i <= nb; i++)
		{
			int temp = 0;
			scanf("%d", &temp);
			if(temp) B += fib[i];
		}
		cin >> nc;
		for(int i = 1; i <= nc; i++)
		{
			int temp = 0;
			scanf("%d", &temp);
			if(temp) C += fib[i];
		}
		k = A * B - C;
		for(int i = 1; i <= 2000000; i++)
		{
			if(fib[i] == k)
			{
				cout << i << endl;
				break;
			}
		}
	}
	return 0;
 }