HihoCoder - 1509

HihoCoder - 1509

Given a sequence a[1..n], you need to calculate how many integers S satisfy the following conditions:

(1). 0 ≤ S < 2^60

(2). For every i in [1,n-1] , (a[i] xor S) ≤ (a[i+1] xor S)

Input

On the first line there is only one integer n

On the second line there are n integers a[1..n]

1 ≤ n ≤ 50

0 ≤ a[i] < 2^60

Output

Output one integer : the answer

题意

对于一个给定的数组，a[i] xor S <= a[i+1] xor S,输出满足条件的S的个数。

思路

对于两个数，假设前面的二进制相同，遇到的第一位不同，1/0，那么S在这一位上的二进制数是固定的，同理我们可以固定其他的以为位置，S一共60位，则S的可能取值的数目就是 2^(60-固定数目)。

#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
ll a[100];
int b1[100], b2[100];
int v[100];
ll ans;
void dl(ll x, ll y) {
    int i, j;
    for (int i = 1; i <= 60; i++) {
        b1[i] = x % 2;
        x = x / 2;
    }
    for (int j = 1; j <= 60; j++) {
        b2[j] = y % 2;
        y = y / 2;
    }
     ans = 0;
    for (int i = 60; i >= 1; i--) {
        if (b1[i] == b2[i])
        continue;
        ans = i;
        break;
    }
    v[ans] = 1;
}

int main() {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++)
    cin >> a[i];
    for (int i = 0; i < n - 1; i++)
    dl(a[i], a[i + 1]);
    ans = 0;
    for (int i = 60; i >= 1; i--) 
    {
        if (v[i])continue;
        ans++;
    }
    ans = (ll)1 << ans;
    cout << ans << endl;
}