Akuna Capital笔试第二题， 找出异或大于与操作的对数量

题意很简单，就是给一个数组，找出有多少个对是x ^ y > x & y

首先这题cf味道很冲，感觉当个div2b不是问题，那么我们就来考虑下高位讨论

 所以我们可以计算出来每个ai的高位是多少，然后等差数列求和一下就行，最后用总对数减掉高位相同的对数，就是我们要的答案

#include <bits/stdc++.h>
using namespace std;
#define limit (1000000 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-9
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a, b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a  ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\akioi\\dabiao.txt", "wt", stdout)
typedef long long ll;
typedef unsigned long long ull;
char buf[1 << 23], *p1 = buf, *p2 = buf, obuf[1 << 23], *O = obuf;

inline ll read() {
#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++)
    ll sign = 1, x = 0;
    char s = getchar();
    while (s > '9' || s < '0') {
        if (s == '-')sign = -1;
        s = getchar();
    }
    while (s >= '0' && s <= '9') {
        x = (x << 3) + (x << 1) + s - '0';
        s = getchar();
    }
    return x * sign;
#undef getchar
}//快读
void print(ll x) {
    if (x / 10) print(x / 10);
    *O++ = x % 10 + '0';
}

void write(ll x, char c = 't') {
    if (x < 0)putchar('-'), x = -x;
    print(x);
    if (!isalpha(c))*O++ = c;
    fwrite(obuf, O - obuf, 1, stdout);
    O = obuf;
}


int n, m, k;
int a[limit];
void solve() {
    cin>>n;
    rep(i,1,n){
        cin>>a[i];
    }
    int bits[32] = {0};
    rep(i,1,n){
        per(j, 0, 31){
            int bit = 1<<j;
            if(a[i] & bit){
                bits[j]++;
                break;
            }
        }
    }
    ll ans = 0;
    ll res = 0;
    rep(i,0,30){
        res += bits[i] * (bits[i] - 1) >> 1;
    }
    ans = n * (n - 1) >> 1;
    cout<<ans - res<<endl;
}
int32_t main() {
#ifdef LOCAL
    FOPEN;
//    FOUT;
#endif
    FASTIO

//    int kase;
//    cin >> kase;
//    while (kase--)
        solve();
    cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << "s\n";
    return 0;
}

标签：typedef,cout,int,rep,long,异或,Capital,Akuna,define
来源： https://www.cnblogs.com/tiany7/p/16462913.html