2021牛客多校8 D OR

Posted 2021-09-05 布图

题目

时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld

题目描述

There are two sequences of length n-1n−1, b=(b_2,b_3,\\ldots,b_n)b=(b2,b3,…,b**n), c=(c_2,c_3,\\ldots,c_n)c=(c2,c3,…,c**n). Here, each b_ib**i,c_ic**i is a non-negative integer.

Now, the sequence a=(a_1,a_2,\\ldots,a_n)a=(a1,a2,…,a**n) considers beautiful if and only if forall ii (2 \\leq i \\leq n)(2≤i≤n), b_i = a_{i-1} , \\text{or} , a_{i}b**i=a**i−1ora**i, c_i = a_{i-1} + a_{i}c**i=a**i−1+a**i and each a_ia**i is a non-negative integer.

Now, Toilet-Ares asks you to calculate the number of beautiful sequences.

输入描述:

The first line contains one integer nn (2 \\leq n \\leq 10^5)(2≤n≤105) - the length of sequence aa.

The second line contains n-1n−1 integers b_2,b_3,\\ldots,b_nb2,b3,…,bn (0 \\leq b_i < 2^{30})(0≤bi<230) - the elements of sequence bb.

The third line contains n-1n−1 integers c_2,c_3,\\ldots,c_nc2,c3,…,cn (0 \\leq c_i < 2^{31})(0≤ci<231) - the elements of sequence cc.

输出描述:

Print one number - the number of beautiful sequences.

示例1

输入

[复制](javascript:void(0)😉

4
7 5 5
7 9 5

输出

[复制](javascript:void(0)😉

2

代码与解释

b[ i ] = a[ i ] | a[ i − 1 ]
c[ i ] = a[ i ] + a[ i − 1 ]
给出n个数字作为b数组,n个为c数组,问这样的a数组有多少个

首先,要讲一个定理
a+b == a|b + a&b

a+b == a|b + a&b
c[i] = b[i] + a[i-1] & a[i]
a[i-1]+a[i]=a[i-1] | a[i] + a[i-1] & a[i]
c[i] - b[i] = a[i-1] & a[i]

这样,外面可以得到a&b的数组,外面把它定义为d数组

由于c数组的存在,我们只需要知道第一个数字a[1],其他数字都都知道了,所以我们只需要把a[1]的每一位遍历一遍即可

剩下的看代码吧

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <iostream>
#include <sstream>
#include <set>
#include <map>
#include <queue>
#include <bitset>
#include <vector>
#include <limits.h>
#include <assert.h>
#include <functional>
#include <numeric>
#include <ctime>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
#define pb          push_back
#define ppb         pop_back
#define lbnd        lower_bound
#define ubnd        upper_bound
#define endl        '\\n'
#define mll         map<ll,ll>
#define msl         map<string,ll>
#define mls         map<ll, string>
#define rep(i,a,b)  for(ll i=a;i<b;i++)
#define repr(i,a,b) for(ll i=b-1;i>=a;i--)
#define trav(a, x)  for(auto& a : x)
#define pll         pair<ll,ll>
#define vl          vector<ll>
#define vll         vector<pair<ll, ll>>
#define vs          vector<string>
#define all(a)      (a).begin(),(a).end()
#define F           first
#define S           second
#define sz(x)       (ll)x.size()
#define hell        1000000007
#define DEBUG       cerr<<"/n>>>I'm Here<<</n"<<endl;
#define display(x)  trav(a,x) cout<<a<<" ";cout<<endl;
#define what_is(x)  cerr << #x << " is " << x << endl;
#define ini(a)      memset(a,0,sizeof(a))
#define ini2(a,b)   memset(a,b,sizeof(a))
#define rep(i,a,b)  for(int i=a;i<=b;i++)
#define case        ll T;cin>>T;for(ll Q=1;Q<=T;Q++)
#define lowbit(x)   x&(-x)
#define pr          printf
#define sc          scanf
#define _           0
#define ordered_set tree<ll, null_type,less<ll>, rb_tree_tag,tree_order_statistics_node_update>
#define FAST ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define DBG(x) \\
    (void)(cout << "L" << __LINE__ \\
    << ": " << #x << " = " << (x) << '\\n')
#define TIE \\
    cin.tie(0);cout.tie(0);\\
    ios::sync_with_stdio(false);
//#define long long int

//using namespace __gnu_pbds;

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const double PI    = acos(-1.0);
const double eps   = 1e-6;
const int    INF   = 0x3f3f3f3f;
const ll     LLINF = 0x3f3f3f3f3f3f3f3f;
const int    maxn  = 1e5+10;
const ll     N     = 5;

ll n, ans = 1, b[maxn], c[maxn], d[maxn];

int check(int w, int x) {
	int OR, AND;
	for (int i=2; i<=n; i++) {
		OR = (b[i]>>w) & 1;//or的第w+1位(因为w开始是0),AND同理
		AND = (d[i]>>w) & 1;
		if (OR && !AND) x^=1;//如果是奇数^=1就是-1,如果是偶数^=1就是+1,相当于最后一位的取反,其他位不变
		else if (!OR && AND) return 0;
		else if (!OR && !AND && x==1) return 0;
		else if (OR && AND && x==0) return 0;
		//解释一下以上四行if else (如果还不懂自己个表格看一看)
		//如果OR=1,AND=0,那么有两种可能,0和1
		//如果OR=0,AND=1,那么完全不可能出现
		//如果OR=0,AND=0,那么只可能x只能是0
		//如果OR=1,AND=1,那么只可能x只能是1
	}
	return 1;
}
	
void solve(){
	cin>>n;
	for (ll i=2; i<=n; i++) cin>>b[i];
	for (ll i=2; i<=n; i++) cin>>c[i], d[i] = c[i] - b[i];
	
	for (int i=0; i<32; i++) {//遍历所有位
		int a = check(i, 0);//这一位是0的可能
		int b = check(i, 1);//这一位是1的可能
		if (a && b) ans*=2;//如果都可以那就乘2
		else if (!a && !b) {//不行就输出0
			cout<<0<<endl;
			return ;
		}
		//其他情况ans不变
	}
	cout<<ans<<endl;
	
}

int main()
{
//	TIE;
//    #ifndef ONLINE_JUDGE
//    freopen ("input.txt","r",stdin);
//    #else
//    #endif
	solve();
//    case{solve();}
//    case{cout<<"Case "<<Q<<":"<<endl;solve();}
	return ~~(0^_^0);
}

参考:https://blog.csdn.net/yanweiqi1754989931/article/details/119542804?ops_request_misc=%257B%2522request%255Fid%2522%253A%2522162884136216780357248510%2522%252C%2522scm%2522%253A%252220140713.130102334.pc%255Fblog.%2522%257D&request_id=162884136216780357248510&biz_id=0&utm_medium=distribute.pc_search_result.none-task-blog-2_blogfirst_rank_v2~rank_v29-11-119542804.pc_v2_rank_blog_default&utm_term=%E7%89%9B%E5%AE%A2%E5%A4%9A%E6%A0%A18&spm=1018.2226.3001.4450