## The Oculus HDU - 6768（ull自然取余溢出 哈希）

2020/8/11 19:13:31 文章标签:

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<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=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<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 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

**思路：**一开始比赛看到这个题，无从下手，肯定会爆数据范围，后来发现，ull自然溢出取余，哈希思想直接暴力就可以做出来，草率了。。。不过也让我知道了原来这就是哈希思想。

``````#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cctype>
#include<iomanip>
#include<map>
#include<vector>
#include<list>
#include<deque>
#include<stack>
#include<queue>
#include<set>
#include<cctype>
#include<string>
#include<stdexcept>
#include<fstream>
#include<sstream>
#define mem(a,b) memset(a,b,sizeof(a))
#define debug() puts("what the fuck!")
#define dedebug() puts("what the fuck!!!")
#define ll long long
#define ull unsigned long long
#define speed {ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); };
using namespace std;
const double PI = acos(-1.0);
const int maxn = 2e6 + 50;
const int N = 110;
const int INF = 0x3f3f3f3f;
const int inf = 0xfffffff;//比INF小，防止累加爆int
const double esp_0 = 1e-6;
const double gold = (1 + sqrt(5)) / 2;
const int mod = 193;
int gcd(int x, int y) {
return y ? gcd(y, x % y) : x;
}
struct node {
ll a, b, c, d;
node(ll a, ll b, ll c, ll d) :a(a), b(b), c(c), d(d) {};
};
ull f[maxn];
void init() {
f = 1;
f = 2;
for (int i = 3; i < maxn; ++i) {
f[i] = f[i - 1] + f[i - 2];
}
}
int main() {
init();
int t;
scanf("%d", &t);
while (t--) {
int a, b, c;
ull A = 0, B = 0, C = 0;
scanf("%d", &a);
for (int i = 1; i <= a; ++i) {
int x;
scanf("%d", &x);
if (x)A += f[i];
}
scanf("%d", &b);
for (int i = 1; i <= b; ++i) {
int x;
scanf("%d", &x);
if (x)B += f[i];
}
scanf("%d", &c);
for (int i = 1; i <= c; ++i) {
int x;
scanf("%d", &x);
if (x)C += f[i];
}
A *= B;
int step = 1;
while (C + f[step] != A) {
step++;
}
printf("%d\n", step);
}
return 0;
}
``````

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