CodeForces 1285D Dr. Evil Underscores（二进制位运算）

Description：

Today, as a friendship gift, Bakry gave Badawy $n$ integers $a_{1},a_{2},…,a_{n}$ and challenged him to choose $a_{n}$ integer $X$ such that the value $max_{1}^{n}(a_{i}⊕X)$ is minimum possible, where ⊕ denotes the bitwise XOR operation.

As always, Badawy is too lazy, so you decided to help him and find the minimum possible value of $max_{1}^{n}(a_{i}⊕X)$ .

Input

The first line contains integer $n (1≤n≤10^5)$ .

The second line contains n integers $a_{1},a_{2},…,a_{n} (0≤a_{i}≤2^{30}−1)$ .

Output

Print one integer — the minimum possible value of $max_{1}^{n}(a_{i}⊕X)$ .

Examples

inputCopy

3
1 2 3

outputCopy

inputCopy

2
1 5

outputCopy

Note

In the first sample, we can choose $X=3$ .

In the second sample, we can choose $X=5$ .

题意：

给你n个数，让你寻找一个X，使X异或这n个数的最大值尽可能的要小。

把每个数据都按二进制去思考，如果这些数某个二进制位置上的数既有0又有1，那么不论取值X的二进制位置上对应的位置上是0还是1，这个位置上异或后的值一定是1！

所以，我们每次统计一下，如果这个位置既有0又有1，那么这个位置异或后肯定是1，如果这个位置上的数据相同，那么我们就不用管它，让这个位置上的值为0，就好了最大值尽可能的小；然后我们这样每次按位处理，最多30次就可以处理完！

AC代码：

#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <queue>
using namespace std;
#define sd(n) scanf("%d", &n)
#define sdd(n, m) scanf("%d%d", &n, &m)
#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define pd(n) printf("%d\n", n)
#define pc(n) printf("%c", n)
#define pdd(n, m) printf("%d %d", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n, m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld", &n)
#define sldd(n, m) scanf("%lld%lld", &n, &m)
#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)
#define sf(n) scanf("%lf", &n)
#define sc(n) scanf("%c", &n)
#define sff(n, m) scanf("%lf%lf", &n, &m)
#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define ss(str) scanf("%s", str)
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define mem(a, n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mod(x) ((x) % MOD)
#define gcd(a, b) __gcd(a, b)
#define lowbit(x) (x & -x)
typedef pair<int, int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
inline int read()
{
    int ret = 0, sgn = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9')
    {
        if (ch == '-')
            sgn = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
    {
        ret = ret * 10 + ch - '0';
        ch = getchar();
    }
    return ret * sgn;
}
inline void Out(int a) //Êä³öÍâ¹Ò
{
    if (a > 9)
        Out(a / 10);
    putchar(a % 10 + '0');
}

ll gcd(ll a, ll b)
{
    return b == 0 ? a : gcd(b, a % b);
}

ll lcm(ll a, ll b)
{
    return a * b / gcd(a, b);
}
///快速幂m^k%mod
ll qpow(ll a, ll b, ll mod)
{
    if (a >= mod)
        a = a % mod + mod;
    ll ans = 1;
    while (b)
    {
        if (b & 1)
        {
            ans = ans * a;
            if (ans >= mod)
                ans = ans % mod + mod;
        }
        a *= a;
        if (a >= mod)
            a = a % mod + mod;
        b >>= 1;
    }
    return ans;
}

// 快速幂求逆元
int Fermat(int a, int p) //费马求a关于b的逆元
{
    return qpow(a, p - 2, p);
}

///扩展欧几里得
int exgcd(int a, int b, int &x, int &y)
{
    if (b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    int g = exgcd(b, a % b, x, y);
    int t = x;
    x = y;
    y = t - a / b * y;
    return g;
}

///使用ecgcd求a的逆元x
int mod_reverse(int a, int p)
{
    int d, x, y;
    d = exgcd(a, p, x, y);
    if (d == 1)
        return (x % p + p) % p;
    else
        return -1;
}

///中国剩余定理模板0
ll china(int a[], int b[], int n) //a[]为除数，b[]为余数
{
    int M = 1, y, x = 0;
    for (int i = 0; i < n; ++i) //算出它们累乘的结果
        M *= a[i];
    for (int i = 0; i < n; ++i)
    {
        int w = M / a[i];
        int tx = 0;
        int t = exgcd(w, a[i], tx, y); //计算逆元
        x = (x + w * (b[i] / t) * x) % M;
    }
    return (x + M) % M;
}

int n;
int a[100010];
int x;
int solve(vector<int> v, int x)
{
    if (x == -1)
        return 0;
    if (v.size() == 0)
        return 0;     //递归到头了，没有数据了；直接返回就可以
    vector<int> l, r; //区分给位置的数值
    for (int i : v)
    {
        if (i & (1 << x))
            l.pb(i); //如果是 1 就放进 l 里面
        else
            r.pb(i); //0就放进 r里面；
    }
    if (l.size() == 0)
        return solve(r, x - 1); //左边为空说明这个位置都是 1 ，直接处理右边
    if (r.size() == 0)
        return solve(l, x - 1);                              //右边为空说明这个位置都是 0 ，直接处理左边
    return min(solve(r, x - 1), solve(l, x - 1)) + (1 << x); //因为是要最小值，所以让最小的加上
} //二进制该位置唯一的十进制数（类似进制转换）

int main()
{
    sd(n);
    vector<int> v;
    for (int i = 0; i < n; i++)
    {
        sd(x);
        v.pb(x);
    }
    int ans = solve(v, 29);
    pd(ans);
    return 0;
}

邵光亮

发布了611 篇原创文章 · 获赞 390 · 访问量 20万+

他的留言板关注