【进制转换＋异或运算】XOR Sorting HihoCoder - 1509

Think:
1知识点：【进制转换＋异或运算】
2题意：输入一个序列a[1..n]，判断满足一下两个条件的S有多少个
(1):0 <= S < pow(2, 60)
(2):对于i(1 <= i < n)，(a[i] xor S) <= (a[i+1] xor S)
3解题思路：
(1):先将a[i]和a[i+1]转换为二进制，然后通过异或运算相同为零不同为一的性质和(a[i] xor S) <= (a[i+1] xor S)的条件确定可以确定的位，然后对于无法确定的位则均有两种取值(0、1)，进而得到满足条件的S的个数

vjudge题目链接

以下为Accepted代码

#include 
#include 
#include 

using namespace std;

typedef long long LL;

int rec[64], s1[64], s2[64];

bool solve(LL x, LL y);
LL power(LL x, int k);

int main(){
    int n, i, cnt, flag;
    LL a, t;
    scanf("%d", &n);
    flag = 1;
    memset(rec, -1, sizeof(rec));
    scanf("%lld", &t);
    for(i = 1; i < n; i++){
        scanf("%lld", &a);
        if(!solve(a, t)) flag = false;
        t = a;
    }
    if(!flag){
        printf("0\n");
    }
    else {
        cnt = 0;
        for(i = 0; i < 60; i++){
            if(rec[i] == -1) cnt++;
        }
        printf("%lld\n", power(2, cnt));
    }
    return 0;
}
LL power(LL x, int k){
    LL ans = 1;
    while(k){
        if(k & 1) ans *= x;
        x *= x;
        k >>= 1;
    }
    return ans;
}
bool solve(LL x, LL y){
    int i;
    for(i = 0; i < 60; i++){
        if(x & ((LL)1 << i)) s1[i] = 1;
        else s1[i] = 0;
        if(y & ((LL)1 << i)) s2[i] = 1;
        else s2[i] = 0;
    }
    for(i = 59; i >= 0; i--){
        if(s1[i] == 1 && s2[i] == 0){
            if(rec[i] == -1 || rec[i] == 1) rec[i] = 1;
            else return false;
            break;
        }
        else if(s1[i] == 0 && s2[i] == 1){
            if(rec[i] == -1 || rec[i] == 0) rec[i] = 0;
            else return false;
            break;
        }
    }
    return true;
}