CodeForces 742B Arpa’s obvious problem and Mehrdad’s terrible solution (暴力枚举)

题意：求定 n 个数，求有多少对数满足，ai^bi = x。

析：暴力枚举就行，n的复杂度。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const double inf = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 1e5 + 5;

const int mod = 1e9 + 7;

const int dr[] = {-1, 1, 0, 0};

const int dc[] = {0, 0, 1, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline int Min(int a, int b){ return a < b ? a : b; }

inline int Max(int a, int b){ return a > b ? a : b; }

inline LL Min(LL a, LL b){ return a < b ? a : b; }

inline LL Max(LL a, LL b){ return a > b ? a : b; }

inline bool is_in(int r, int c){

    return r >= 0 && r < n && c >= 0 && c < m;

}

int a[100005];

int b[100005];

int main(){

    LL n, m;

    while(scanf("%I64d %I64d", &n, &m) == 2){

        for(int i = 0; i < n; ++i)  scanf("%d", a+i);

        memset(b, 0, sizeof b);

        LL ans = 0;

//        if(m == 0){

//            for()

//            continue;

//        }

        for(int i = n-1; i >= 0; --i){

            LL t = a[i] ^ m;

            if(t < 100005)  ans += b[t];

            ++b[a[i]];

        }

        cout << ans << endl;

    }

    return 0;

}