2020杭电多校第二场The Oculus

第六题：The Oculus

题目：

Let’s define the Fibonacci sequence F1,F2,… as F1=1,F2=2,Fi=Fi−1+Fi−2 (i≥3).

It’s well known that every positive integer x has its unique Fibonacci representation (b1,b2,…,bn) such that:

· b1×F1+b2×F2+⋯+bn×Fn=x.

· bn=1, and for each i (1≤i

· For each i (1≤i

For example, 4=(1,0,1), 5=(0,0,0,1), and 20=(0,1,0,1,0,1) because 20=F2+F4+F6=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 (b1,b2,…,bn), Little Q then selected a bit k (1≤k

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.

思路

双哈希

代码

#include
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;


const int N=2e6+9;
const int mod=1e9+7;
const int mod1=1e9+9;
ll c[N];
ll fib[N],fib1[N];

int main()
{
	fib[1]=1,fib[2]=2;
	fib1[1]=1,fib1[2]=2;
	for(int i=3;i<N;i++)
	{
		fib[i]=(fib[i-1]+fib[i-2])%mod;
		fib1[i]=(fib1[i-1]+fib1[i-2])%mod1; 
	}
	int T;
	cin>>T;
	while(T--)
	{
		int n,m,t;
		scanf("%d",&n);
		ll ans1=0,ans11=0,ans2=0,ans22=0,ans3=0,ans33=0;
		for(int i=1;i<=n;i++)
		{
			ll x;
			scanf("%lld",&x);
			if(x)
			{
				ans1=(ans1+fib[i])%mod;
				ans11=(ans11+fib1[i])%mod1;
			}
		}
		scanf("%d",&m);
		for(int i=1;i<=m;i++)
		{
			ll x;
			scanf("%lld",&x);
			if(x)
			{
				ans2=(ans2+fib[i])%mod;
				ans22=(ans22+fib1[i])%mod1;
			}
		}
		scanf("%d",&t);
		for(int i=1;i<=t;i++)
		{
			scanf("%lld",&c[i]);
			if(c[i])
			{
				ans3=(ans3+fib[i])%mod;
				ans33=(ans33+fib1[i])%mod1;
			}
		}
		ans2=(ans2*ans1)%mod;
		ans22=(ans22*ans11)%mod1;
		
		
		for(int i=1;i<=t;i++)
		{
			ll ans=ans3,anss=ans33;
			
			if(!c[i])
			{
				ans=(ans+fib[i])%mod;
				anss=(anss+fib1[i])%mod1;
				if(ans2==ans&&ans22==anss)
				{
					printf("%d\n",i);
					break;
				}
			}
			
		}
	}
	return 0;
}