300iq Contest 3

B. Best Tree

首先n==2使只有一对；
n!=2时，如果1的个数比别的个数多，即别的点都和1匹配
如果1个个数比别的个数少，所有点即可俩俩匹配，

#include <bits/stdc++.h>

#define int long long
using namespace std;
const int mod = 998244353;
const int maxn = 1e6 + 10;

void solve() {
    int n,x;
    cin>>n;
    int none=0;
    for (int i = 1; i <=n; ++i) {
        cin>>x;
        if(x!=1) none++;
    }
    if(n==2) cout<<1<<endl;
    else cout<<min(n/2,none)<<endl;
}

signed main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int _ = 1;
    cin >> _;
    while (_--)
        solve();
    return 0;
}

I. Ignore Submasks

俩个数&只要一个数的2进制的某一个1为0即可使&的结果不一样
遍历数组，记录使用过的二进制位置即可

#include <bits/stdc++.h>

#define int long long
using namespace std;
const int mod = 998244353;
const int maxn = 1e6 + 10;

int a[105];
int vis[70];

int fastpow(int a, int n) {
    if (n < 0) return 0;
    int res = 1, temp = a;
    while (n) {
        if (n & 1) res = res * temp % mod;
        temp = temp * temp % mod;
        n >>= 1;
    }
    return res % mod;
}

void solve() {
    memset(vis, 0, sizeof(vis));
    int n, k, kk = 0, sum = 0;
    cin >> n >> k;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    for (int i = 1; i <= n; ++i) {
        int x = a[i];
        int aa = 0;
        for (int j = 0; j < k; ++j) {
            if (vis[j] == 1)continue;
            if (x & (1ll << j)) {
                aa++;
                vis[j] = 1;
                kk++;
            }
        }
        int b = k - kk;
        sum = (sum + (fastpow(2, aa) - 1) * fastpow(2, b) % mod * i % mod) % mod;
    }
    cout << sum % mod << endl;
}

signed main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int _ = 1;
//    cin >> _;
    while (_--)
        solve();
    return 0;
}

本文地址：https://blog.csdn.net/weixin_45436102/article/details/110644888