Codeforces Round #439 (Div. 2) 题解
题目链接 Round 439 div2
就做了两道题TAT
开场看C题就不会
然后想了好久才想到。
三种颜色挑出两种算方案数其实是独立的,于是就可以乘起来了。
E题想了一会有了思路,然后YY出了一种方案。
我们可以对每个矩形随机一个权值,然后用二维树状数组搞下。
询问的时候看两个点权值是否相等就可以了
于是就过了。
D题待补
给出一棵完全二叉树,这棵树上有附带的m条边(m <= 4),求这张图的简单路径条数。
qls的题就是厉害……
C题
#include <bits/stdc++.h> using namespace std; #define rep(i, a, b) for (int i(a); i <= (b); ++i) #define dec(i, a, b) for (int i(a); i >= (b); --i) #define MP make_pair #define fi first #define se second typedef long long LL; const int N = 5010; const int mod = 998244353; int P[N][N], C[N][N]; int a, b, c; int calc(int x, int y){ if (x > y) swap(x, y); LL ret = 0; rep(i, 0, x) ret = (ret + 1ll * P[y][i] * C[x][i]) % mod; return ret; } int main(){ C[0][0] = P[0][0] = 1; scanf("%d%d%d", &a, &b, &c); rep(i, 1, 5000){ C[i][0] = P[i][0] = 1; rep(j, 1, i){ P[i][j] = (1ll * P[i - 1][j - 1] * j % mod + 1ll * P[i - 1][j] % mod) % mod; C[i][j] = (1ll * C[i - 1][j - 1] + 1ll * C[i - 1][j]) % mod; } } int ans = 1ll * calc(a, b) % mod * 1ll * calc(b, c) % mod * 1ll * calc(a, c) % mod; printf("%d\n", ans); return 0; }
E题
#include <bits/stdc++.h> using namespace std; #define rep(i, a, b) for (int i(a); i <= (b); ++i) #define dec(i, a, b) for (int i(a); i >= (b); --i) #define MP make_pair typedef long long LL; typedef pair <int, int> PII; const LL mod = 1e9 + 7; const int N = 5010; map <PII, LL> mp; int cnt = 0; int n, m, q; int p[N][N]; LL c[N][N]; inline void add(int x, int y, LL val){ for (; x <= n; x += x & -x){ for (int t = y; t <= m; t += t & -t){ c[x][t] = (c[x][t] + val) % mod; c[x][t] = (c[x][t] + mod) % mod; } } } inline LL query(int x, int y){ LL ret = 0; for (; x ; x -= x & -x){ for (int t = y; t ; t -= t & -t){ (ret += c[x][t]) %= mod; } } return ret; } inline LL solve(int x, int y){ LL ret = query(x, y) % mod; return ret; } int main(){ scanf("%d%d%d", &n, &m, &q); rep(i, 1, n) rep(j, 1, m) p[i][j] = ++cnt; rep(i, 1, q){ int op; scanf("%d", &op); if (op == 1){ int x1, y1, x2, y2; scanf("%d%d%d%d", &x1, &y1, &x2, &y2); if (x1 > x2) swap(x1, x2); if (y1 > y2) swap(y1, y2); LL cnt = (LL)rand() * (LL)rand() * (LL)rand() % mod; mp[MP(p[x1][y1], p[x2][y2])] = cnt; add(x1, y1, cnt); add(x1, y2 + 1, -cnt); add(x2 + 1, y1, -cnt); add(x2 + 1, y2 + 1, cnt); } else if (op == 2){ int x1, y1, x2, y2; scanf("%d%d%d%d", &x1, &y1, &x2, &y2); if (x1 > x2) swap(x1, x2); if (y1 > y2) swap(y1, y2); LL cnt = mp[MP(p[x1][y1], p[x2][y2])]; add(x1, y1, -cnt); add(x1, y2 + 1, cnt); add(x2 + 1, y1, cnt); add(x2 + 1, y2 + 1, -cnt); } else{ int x1, y1, x2, y2; scanf("%d%d%d%d", &x1, &y1, &x2, &y2); LL xx = solve(x1, y1); LL yy = solve(x2, y2); if (xx == yy) puts("Yes"); else puts("No"); } } return 0; }