Codeforces Round #439 (Div. 2)

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A. The Artful Expedient

由于xi⊕yj=xk⟺xk⊕yj=xi，所以一定是偶数个。

Karen

B. The Eternal Immortality

差超过5的一定是0，否则暴力就好。

#include <bits/stdc++.h>using namespace std;long long a, b;int main(){    cin >> a >> b;    if (b-a > 20) puts("0");    else {        long long ans = 1;        for (long long i = b; i > a; i--)            ans = ans*(i%10)%10;        cout << ans << endl;    }    return 0;}

C. The Intriguing Obsession

三种颜色的岛屿分别有a,b,c个，要求连边使得同色两点间距离至少是3，问如何连接。

显然这个要求等价于不能有一个节点连向两个同色节点。考虑分别由a向b，a向c，c向b连边。a向b连边为例，则连边方案数是：

∑i≤a(ai)(bi)i!

暴力计算即可。

#include <bits/stdc++.h>using namespace std;const int MAXN = 5005, mod = 998244353;int C[MAXN][MAXN], fac[MAXN];int a, b, c;int main(){    scanf("%d%d%d", &a, &b, &c);    C[0][0] = 1;    for (int i = 1; i <= 5000; i++)        for (int j = 0; j <= i; j++)            C[i][j] = (C[i-1][j]+C[i-1][j-1]*(j!=0))%mod;    fac[0] = 1;    for (int i = 1; i <= 5000; i++)        fac[i] = (long long)fac[i-1]*i%mod;    int ans[3] = {0, 0, 0};    for (int i = 0; i <= a; i++)         ans[0] = (ans[0]+(long long)C[a][i]*C[b][i]%mod*fac[i]%mod)%mod;    for (int i = 0; i <= a; i++)        ans[1] = (ans[1]+(long long)C[a][i]*C[c][i]%mod*fac[i]%mod)%mod;    for (int i = 0; i <= c; i++)        ans[2] = (ans[2]+(long long)C[c][i]*C[b][i]%mod*fac[i]%mod)%mod;    cout << (long long)ans[0]*ans[1]%mod*ans[2]%mod << endl;    return 0;}

E. The Untended Antiquity(补)

题意：一个n×m的矩形，多次操作：

1. 在一个矩形外安栅栏
2. 拆掉一个安过的栅栏
3. 询问两点是否联通

没见过的套路orz，就是给每个栅栏随机一个权值，然后用二维树状数组大力维护一下，每次安栅栏就给矩形内的点异或一个权值。如果两点权值相等则两点联通。

#include <bits/stdc++.h>using namespace std;const int MAXN = 2505;typedef unsigned long long ull;ull c[MAXN][MAXN];int n, m, q;inline int lowbit(int nd){ return nd&(-nd); }struct pt {    int x1, y1, x2, y2;    friend bool operator < (const pt &a, const pt &b)    {        if (a.x1 != b.x1) return a.x1 < b.x1;        if (a.y1 != b.y1) return a.y1 < b.y1;        if (a.x2 != b.x2) return a.x2 < b.x2;        return a.y2 < b.y2;    }};map<pt, ull> dt;void modify(int x, int y, ull dt){    for (int i = x; i <= n; i += lowbit(i))        for (int j = y; j <= m; j += lowbit(j))            c[i][j] ^= dt;}ull sum(int x, int y){    ull ans = 0;    for (int i = x; i; i -= lowbit(i))        for (int j = y; j; j -= lowbit(j))            ans ^= c[i][j];    return ans;}void cover(int x1, int y1, int x2, int y2, ull dt){    modify(x1, y1, dt);    modify(x1, y2+1, dt);    modify(x2+1, y1, dt);    modify(x2+1, y2+1, dt);}int main(){    srand(time(0));    scanf("%d%d%d", &n, &m, &q);    for (int i = 1; i <= q; i++) {        int t, r1, r2, c1, c2;        scanf("%d%d%d%d%d", &t, &r1, &c1, &r2, &c2);        if (t == 1) {            ull rk = rand()*RAND_MAX+rand();            // cerr << rk << endl;            dt[(pt){r1, c1, r2, c2}] = rk;            cover(r1, c1, r2, c2, rk);        } else if (t == 2) {            ull rk = dt[(pt){r1, c1, r2, c2}];            cover(r1, c1, r2, c2, rk);        } else {            if (sum(r1, c1) == sum(r2, c2)) puts("Yes");            else puts("No");        }    }    return 0;}