2020杭电HDU多校第二场The Oculus（假斐波那契数列真Hash）

Problem Description

Let's define the Fibonacci sequence \(F_1,F_2,… as F_1=1,F_2=2,F_i=F_{i−1}+F_{i−2}\) (i≥3).

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

· \(b_1×F_1+b_2×F_2+⋯+b_n×F_n=x\).

· bn=1, and for each i (1≤i<n), bi∈{0,1} always holds.

· For each i (1≤i<n), bi×bi+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 \((b_1,b_2,…,b_n)\), Little Q then selected a bit k (1≤k<n) such that bk=1 and modified bk 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≤T≤10000), 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 \(b_1,b_2,…,b_n\), denoting the value of each bit.

It is guaranteed that:

· 1≤|A|,|B|≤1000000.

· 2≤|C|≤|A|+|B|+1.

·∑|A|,∑|B|≤5000000.

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

emmm，题目是真的假。。。看起来唬人的一批，好像还要跑NTT来着。。。结果一个Hash就过去了。。。我们直接暴力计算A和B的值，然后乘一下，跟C的值进行比较就可以了。不过在这之前我们要把斐波那契的Hash值预处理处理。

以下是AC代码：

#include <bits/stdc++.h>
using namespace std;

typedef unsigned long long ull;

const int mac=1e6+10;

ull dp[mac*2];
int a[mac],b[mac],c[mac*2];

int main(int argc, char const *argv[])
{
	int t;
	dp[1]=1;dp[2]=2;
	for (int i=3; i<mac*2; i++)
		dp[i]=dp[i-1]+dp[i-2];
	scanf ("%d",&t);
	while (t--){
		int na,nb,nc;
		scanf ("%d",&na);
		for (int i=1; i<=na; i++)
			scanf ("%d",&a[i]);
		scanf ("%d",&nb);
		for (int i=1; i<=nb; i++)
			scanf ("%d",&b[i]);
		scanf ("%d",&nc);
		for (int i=1; i<=nc; i++)
			scanf ("%d",&c[i]);
		ull s1=0,s2=0,s3=0;
		for (int i=1; i<=na; i++)
			if (a[i]) s1+=dp[i];
		for (int i=1; i<=nb; i++)
			if (b[i]) s2+=dp[i];
		for (int i=1; i<=nc; i++)
			if (c[i]) s3+=dp[i];
		for (int i=1; i<=na+nb+1; i++){
			if (s3+dp[i]==s1*s2){
				printf("%d\n",i);
				break;
			}
		}
	}
	return 0;
}