2020 Multi-University Training Contest 2 The Oculus（Hash）

The Oculus

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1077    Accepted Submission(s): 306

 

Problem Description

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.

 

 

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 b1,b2,…,bn, 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

 

 

Source

2020 Multi-University Training Contest 2

 

 

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题意：

先给出一个序列如上。

给出一个长度为a的01序列表示A，第i个数字为0表示不加F[i]否则加F[i]。

B、C同理，A、B均为真实值，C为A*B，但是数字里面有一位1被改成了0，问你哪一位。

直接hash就可以了，复杂度O(n)。

#include 
using namespace std;
const int N = 2e6 + 20;
const int mod = 1e9 + 9;
typedef long long ll;
typedef unsigned long long ull;
#ifdef LOCAL
#define TIME cout << "RuningTime: " << clock() << "ms\n", 0
#else
#define TIME 0
#endif
ull f[N];
int z[N];
int main()
{
#ifdef LOCAL
	freopen("E:\\input.txt", "r", stdin);
#endif
	f[1] = 1, f[2] = 2;
	for (int i = 3; i <= 2000010; i++)
	{
		f[i] = f[i - 1] + f[i - 2];
	}
	int t;
	cin >> t;
	while (t--)
	{
		int a, b, c;
		scanf("%d", &a);
		ll res1 = 0, res2 = 0, res3 = 0;
		for (int i = 1; i <= a; i++)
		{
			int num;
			scanf("%d", &num);
			res1 = res1 + f[i * num];
		}
		scanf("%d", &b);
		for (int i = 1; i <= b; i++)
		{
			int num;
			scanf("%d", &num);
			res2 = res2 + f[i * num];
		}
		scanf("%d", &c);
		for (int i = 1; i <= c; i++)
		{
			scanf("%d", &z[i]);
			res3 = res3 + f[i * z[i]];
		}
		res1 *= res2;
		for (int i = 1; i <= c; i++)
		{
			if (z[i] == 0)
			{
				ull now = res3 + f[i];
				if (now == res1)
				{
					cout << i << endl;
					break;
				}
			}
		}
	}
	return TIME;
}