CodeForces 742B Arpa’s obvious problem and Mehrdad’s terrible solution

题意： 对于给定的n个数字，要求找到这里面存在有几对数字对pair(a,b)，使得a⊕b=m.

• 发现数字范围是[0,100000]，所以，我们用一个数组cot[300000]，来记录每个数字i出现的次数cot[i]，然后每次加入新的数字x更新ans+=cot[x⊕m]

#include <cstdio>#include <string>#include<iostream>#include<vector>#include <stack>#include <queue>#include <map>#include <cstdlib>#include<string.h>#include <cstring>#include <ctime>#include <algorithm>#include <set>using namespace std;typedef long long ll;typedef pair<int, int>pii;typedef pair<ll, ll> pll;typedef pair<int, ll> pil;typedef vector<vector<ll> >vvi;typedef vector<ll> vi;const ll mod = 1e9 + 7;const int MAXN = 1000;const int MAXM = 1000;ll pow_mod(ll a, ll n, ll mod){    ll res = 1;    while (n)    {        if (n & 1)res = res*a%mod;        a = a*a%mod;        n >>= 1;    }    return res;}ll gcd(ll x, ll y){    return y == 0 ? x : gcd(y, x%y);}ll lcm(ll x, ll y){    return x / gcd(x, y)*y;}struct Edge{    int to; int next;    int cal;}edge[MAXM];int head[MAXN], tot;void addedge(int u, int v, int cal){    edge[tot].to = v;    edge[tot].cal = cal;    edge[tot].next = head[u];    head[u] = tot++;}void init(){    tot = 0;    memset(head, -1, sizeof head);}struct Matrix{    int m[10][10];    int n;    void E()    {        for (int i = 0; i < 10; i++)for (int j = 0; j < 10; j++)            m[i][j] = i == j;    }    void clear()    {        memset(m, 0, sizeof m);    }    Matrix operator*(const Matrix b)    {        Matrix res;        res.n = b.n;        res.clear();        for (int i = 0; i < n; i++)for (int j = 0; j < n; j++)        {            res.m[i][j] = 0;            for (int k = 0; k < n; k++)                res.m[i][j] = (res.m[i][j] + m[i][k] * b.m[k][j] % mod) % mod;        }        return res;    }    Matrix operator^(ll tim)    {        Matrix a = *this;        Matrix res;        res = a;        res.E();        while (tim)        {            if (tim & 1)res = res*a;            a = a*a;            tim >>= 1;        }        return res;    }};int cot[300000 + 10];int main(){    memset(cot, 0, sizeof cot);    int n; int m;    scanf("%d%d", &n, &m);    ll ans = 0;    for (int i = 0; i < n; i++)    {        int x;        scanf("%d", &x);        ll plus = m^x;        ans += (ll)cot[plus];        cot[x]++;    }    printf("%I64d\n", ans);    //getchar();    //getchar();}