Educational Codeforces Round 14

Educational Codeforces Round 14

A Fashion in Berland

做法：模拟
代码：

void solve(){
    int n;
    cin >> n;
    int ans = 0;
    for (int i = 1;i <= n;i ++) {
        int x;
        cin >> x;
        ans += x;
    }  
    if(n == 1) {
        if(ans == 1)
        cout << "YES" << endl;
        else cout << "NO" << endl;
    }else if(n > 1) {
        if(ans == n - 1) cout << "YES" << endl;
        else cout << "NO" << endl;
    }
}

B s-palindrome

做法：模拟，前三题都纯模拟 看代码就好了
代码：

bool check(char a,char b) {
    if(a == b) {
        if(a == 'A' || a == 'H' || a == 'W' || a == 'T' || a == 'Y' || a == 'U' || a == 'I' || a == 'O') return 1;
        if(a == 'M' || a == 'V' || a == 'X') return 1;
        if(a == 'o' || a == 'w' || a == 'x' || a == 'v') return 1;
        return 0;
    }
    else {
        if(a == 'b' && b == 'd') return 1;
        if(a == 'p' && b == 'q') return 1;
        if(a == 'd' && b == 'b') return 1;
        if(a == 'q' && b == 'p') return 1;
        return 0;
    }
    return 0;
}
bool checklast(char a) {
     if(a == 'A' || a == 'H' || a == 'W' || a == 'T' || a == 'Y' || a == 'U' || a == 'I' || a == 'O') return 1;
    if(a == 'M' || a == 'V' || a == 'X') return 1;
    if(a == 'o' || a == 'w' || a == 'x' || a == 'v') return 1;
        return 0;
}
void solve(){   
    string s;
    cin >> s;
    int n = s.size();
    s = " " + s;
    bool f = 1;
    for (int i = 1;i <= n / 2;i ++ ) {
        if(check(s[i] , s[n - i + 1]) == 0) f = 0;
    }
    if(n & 1) {
        if(checklast(s[n / 2 + 1]) == 0) f = 0;
    }
    cout << ((f == 1) ? "TAK": "NIE") << endl;
}

C Exponential notation

做法：纯纯的模拟，有点恶心，要分小数点的前后来看
代码:

void solve(){   
    string s;
    cin >> s;
    int n = s.size();
    s = " " + s;
    int pos = n + 1;
    for (int i = 1;i <= n;i ++) {
        if(s[i] == '.') pos = i;
    }
    if(pos == 0) pos == n + 1;
    int stq = -1,sth = -1; 
    for (int i = 1;i <= n;i ++) {
        if(s[i] != '0' && s[i] != '.') {
            stq = i;
            break;
        }
    }
    for (int i = n;i >= 1;i --) {
        if(s[i] != '0' && s[i] != '.') {
            sth = i;
            break;
        }
    }
    if(stq == -1 && sth == -1) {
        cout << 0 << endl;
        return;
    }
    cout << s[stq];
    if(stq != sth) cout << '.';
    for (int i = stq + 1;i <= sth;i ++) {
        if(s[i] == '.') continue;
        cout << s[i];
    }
    if(stq > pos) {
        if(stq - pos != 0)
        cout << "E-" << stq - pos << endl;
    }
    else {
        if(pos - stq - 1 != 0)
        cout << "E" << pos - stq - 1 << endl;
    }
}

D Swaps in Permutation

做法:这种排列题想到用图去做是一件很显然的思路，所以其实就是用并查集去维护每一个连通块，把每一个连通分块的数放在堆上，按顺序一个个输出就可以了。
代码：

priority_queue<int> q[N];
int f[N];
int find(int x) {
    if(x != f[x]) f[x] = find(f[x]);
    return f[x];
}
void solve(){   
    int n , m;
    cin >> n >> m;
    vector<int> a(n + 1);
    for (int i = 1;i <= n;i ++) cin >> a[i], f[i] = i;
    for (int i = 1;i <= m;i ++) {
        int x, y;
        cin >> x >> y;
        x = find(x);
        y = find(y);
        if(x != y) f[x] = y;
    }
    for (int i = 1;i <= n;i ++) q[find(i)].emplace(a[i]);
    // for (int i = 1;i <= n;i ++) cout << find(i) << ' ';
    for (int i = 1;i <= n;i ++) {
        cout << q[find(i)].top() << ' ';
        q[find(i)].pop();
    }
    cout << endl;
}

E Xor-sequences

做法：很棒的一道矩阵加速加dp题，这道题我一开始做的时候甚至没有想到暴力的递推去做，其实想想是比较显然的，我们可以这么考虑，当前元素是否能放入，那么只和前面一个元素有关，状态可以这么去考虑，\(dp_{i,j}\)为当前放到第\(i\)个位置，放第\(j\)个元素的方案数，那么递推方程为\(dp_{i, j} = \sum_{k = 1}^{n}dp_{i - 1,k}\ast \left [ popcount(a_{k}\bigoplus a_{j} |3) \right ]\)
\(k\)是\(10^{18}\)所以显然这个方程直接去做是超时的，我们可以考虑矩阵加速，因为这个\(popcount\)是可以预处理出来的，做一个矩阵快速幂即可。
代码:

int sz;
struct mat {
	int a[115][115];
	inline mat() { memset(a, 0, sizeof a); }
 
	inline mat operator-(const mat& T) const {
	    mat res;
	    for (int i = 1; i <= sz; ++i)
	      for (int j = 1; j <= sz; ++j) {
	        res.a[i][j] = (a[i][j] - T.a[i][j]) % mod;
	      }
	    return res;
 	 }
 
	inline mat operator+(const mat& T) const {
	    mat res;
	    for (int i = 1; i <= sz; ++i)
	      for (int j = 1; j <= sz; ++j) {
	        res.a[i][j] = (a[i][j] + T.a[i][j]) % mod;
	      }
	    return res;
 	}
 
  	inline mat operator*(const mat& T) const {
	    mat res;
	    int r;
	    for (int i = 1; i <= sz; ++i)
	      for (int k = 1; k <= sz; ++k) {
	        r = a[i][k];
	        for (int j = 1; j <= sz; ++j)
	          res.a[i][j] += T.a[k][j] * r, res.a[i][j] %= mod;
	      }
	    return res;
    }
 
 	inline mat operator^(int x) const {
	    mat res, bas;
	    for (int i = 1; i <= sz; ++i) res.a[i][i] = 1;
	    for (int i = 1; i <= sz; ++i)
	      for (int j = 1; j <= sz; ++j) bas.a[i][j] = a[i][j] % mod;
	    while (x) {
	      if (x & 1) res = res * bas;
	      bas = bas * bas;
	      x >>= 1;
             }
    return res;
    }
}; 
const double eps = 1e-8;
// const int M = N * 4;
mt19937 rng((unsigned int) chrono::steady_clock::now().time_since_epoch().count());
void solve(){   
    int n ,k ;
    cin >> n >> k;
    vector<int> a(n + 1);
    for (int i = 1;i <= n;i ++) cin >> a[i];
    vector<int> f(n + 1);
    mat C;
    sz = n;
    for (int i = 1;i <= n;i ++) {
        for (int j = 1;j <= n;j ++) {
            int x = a[i] ^ a[j];
            int cnt = 0;
            while(x) {
                if(x & 1) cnt++;
                x >>= 1;
            }
            if(cnt % 3 == 0) C.a[i][j] = 1;
        }
    }
    C = C.operator^(k - 1);
    int ans = 0;
    mat Ori;
    for (int i = 1;i <= n;i ++) Ori.a[1][i] = 1;
    for (int i = 1;i <= n;i ++) {
        for (int j = 1;j <= n;j ++) {
            ans = (ans + Ori.a[1][j] * C.a[j][i] % mod) % mod;
        }
    }
    cout << ans << endl;
}