hihoCoder Challenge 28 异或问题思维 - 服务器托管|北京服务器租用|机房托管租用|IDC托管租用|机房机柜带宽租用-价格及费用咨询

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

(1). 0 ≤ S

(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]

Output
Output one integer : the answer

Sample Input
3
1 2 3
Sample Output
288230376151711744

题目大意很明确，思路是什么？
看到异或自然想到二进制，
比如：
00011101
00011010
比较大小从高到低进行比较，如果在某一位置开始出现不同，那么自然异或值S这一位就确定了，因为相同的异或结果还是相同，所以只需要确定哪些位置已经被确定了，那么总方案数就是 2^( 未确定位置数量 ).

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
typedef long long ll;
using namespace std;
typedef unsigned long long int ull;
#define maxn 2000005
#define inf 0x3f3f3f3f
const long long int mod = 1e9 + 7;

ll read() {
    ll x = 0, f = 1;
    char ch = getchar();
    while (ch '9') {
        if (ch == '-') {
            f = -1;
        }
        ch = getchar();
    }
    while (ch >= '0'&&ch  0) {
        if (b % 2)ans = ans * a;
        b = b / 2;
        a = a * a;
    }
    return ans;
}

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


bool cmp(int a, int b) {
    return a > b;
}

ll ves[maxn];
void revs() {
    for (ll i = 1; i = 1; i--) {
        if (b1[i] == b2[i])continue;
        ans = i;
        break;
    }
    v[ans] = 1;
}

int main() {
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    for (int i = 0; i > a[i];
    for (int i = 0; i = 1; i--) {
        if (v[i])continue;
        ans++;
    }
    ans = (ll)1

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