B. So Easy || 思维
行列之差，注意边界处理
#include <bits/stdc++.h> using namespace std; int a[1111][1111]; int main() { int n; cin >> n; int idx, idy; for(int i = 1; i <= n; ++i) { for(int j = 1; j <= n; ++j) { scanf("%d", &a[i][j]); if(a[i][j] == -1) { idx = i; idy = j; } } } //cout << idx << ' ' << idy << endl; if(idx == n) { if(idy == n) cout << a[idx-1][idy] - a[idx-1][idy-1] + a[idx][idy-1] << endl; else cout << a[idx-1][idy] - a[idx-1][idy+1] + a[idx][idy+1] << endl; } else { if(idy == n) cout << a[idx][idy-1] - a[idx+1][idy-1] + a[idx+1][idy] << endl; else cout << a[idx][idy+1] - a[idx+1][idy+1] + a[idx+1][idy] << endl; } }
G. Pot!! || 线段树
维护2、3、5、7四棵线段树，由于代码复用，所以在函数中添加int* tree和int* lazy
#include <bits/stdc++.h> using namespace std; const int maxn = 1e5 + 7; int a[maxn], tr2[maxn<<2], tr3[maxn<<2], tr5[maxn<<2], tr7[maxn<<2], lz2[maxn<<2], lz3[maxn<<2], lz5[maxn<<2], lz7[maxn<<2]; void build(int* tree, int node, int l, int r) { if(l == r) { tree[node] = a[l]; return; } int mid = (l + r) / 2; build(tree, node * 2, l, mid); build(tree, node * 2 + 1, mid + 1, r); tree[node] = max(tree[node*2], tree[node*2+1]); } void push_down(int* tree, int* lazy, int node) { if(lazy[node]) { tree[2*node] += lazy[node]; tree[2*node+1] += lazy[node]; lazy[2*node] += lazy[node]; lazy[2*node+1] += lazy[node]; lazy[node] = 0; } } void update(int* tree, int* lazy, int node, int l, int r, int x, int y, int c) { if(x <= l && y >= r) { tree[node] += c; lazy[node] += c; return; } push_down(tree, lazy, node); int mid = (l + r) / 2; if(x <= mid) update(tree, lazy, node * 2, l, mid, x, y, c); if(y > mid) update(tree, lazy, node * 2 + 1, mid + 1, r, x, y, c); tree[node] = max(tree[node*2], tree[node*2 + 1]); } int query(int* tree, int* lazy, int node, int l, int r, int x, int y) { if(x <= l && y >= r) return tree[node]; push_down(tree, lazy, node); int ans = 0; int mid = (l + r) / 2; if(x <= mid) ans = query(tree, lazy, node * 2, l, mid, x, y); if(y > mid) ans = max(ans, query(tree, lazy, node * 2 + 1, mid + 1, r, x, y)); return ans; } int main() { int n, q, x, y, w; cin >> n >> q; string s; //fill(a + 1, a + 1 + n, 1); 开始不是1！！！！！ 是0！！！！！！ build(tr2, 1, 1, n); build(tr3, 1, 1, n); build(tr5, 1, 1, n); build(tr7, 1, 1, n); while(q--) { cin >> s >> x >> y; if(s == "MULTIPLY") { cin >> w; if(w == 2) update(tr2, lz2, 1, 1, n, x, y, 1); else if(w == 3) update(tr3, lz3, 1, 1, n, x, y, 1); else if(w == 4) update(tr2, lz2, 1, 1, n, x, y, 2); else if(w == 5) update(tr5, lz5, 1, 1, n, x, y, 1); else if(w == 6) { update(tr3, lz3, 1, 1, n, x, y, 1); update(tr2, lz2, 1, 1, n, x, y, 1); } else if(w == 7) update(tr7, lz7, 1, 1, n, x, y, 1); else if(w == 8) update(tr2, lz2, 1, 1, n, x, y, 3); else if(w == 9) update(tr3, lz3, 1, 1, n, x, y, 2); else if(w == 10) { update(tr2, lz2, 1, 1, n, x, y, 1); update(tr5, lz5, 1, 1, n, x, y, 1); } } else { printf("ANSWER %d ", max({query(tr2, lz2, 1, 1, n, x, y), query(tr3, lz3, 1, 1, n, x, y), query(tr5, lz5, 1, 1, n, x, y), query(tr7, lz7, 1, 1, n, x, y)})); //printf("2: %d ", query(tr2, lz2, 1, 1, n, x, y)); //printf("3: %d ", query(tr3, lz3, 1, 1, n, x, y)); //printf("5: %d ", query(tr5, lz5, 1, 1, n, x, y)); //printf("7: %d ", query(tr7, lz7, 1, 1, n, x, y)); } } }
F. Function! || 数学
#include <bits/stdc++.h> using namespace std; typedef long long ll; const int mod = 998244353; ll qpow(ll a, ll b) { ll t = 1; for(; b; b >>= 1, a = a * a % mod) if(b & 1) t = t * a % mod; return t; } ll cal(ll x) { return (x % mod * ((x + 1) % mod) % mod * ((2 * x + 1) % mod) % mod) * qpow(6, mod - 2) % mod; // 相乘每个数都要先取模，因为ll * ll可能会爆ll } int main() { ll n; cin >> n; int x = sqrt(n); ll sum = 0; for(int a = 2; a <= x; ++a) { ll tmp = a; ll cnt = 1; while(tmp <= n) { sum += min(tmp * (a - 1), n - tmp + 1) * cnt * a; //cout << sum << endl; sum %= mod; ++cnt; tmp *= a; } //cout << "tmp " << tmp << endl; //cout << a << endl; //cout << sum << endl; } //cout << sum << endl; //cout << (cal(n) - cal(x)) % mod << endl; sum -= cal(n) - cal(x); //cout << sum << endl; sum += 1LL * (n + 1) % mod * ((x + 1 + n) % mod) % mod * ((n - x) % mod) % mod * qpow(2, mod - 2) % mod;// 相乘每个数都要先取模，因为ll * ll可能会爆ll
cout << (sum % mod + mod) % mod << endl; // 防止sum为负数，且注意上一行求得的sum还未取模，先要取一次模
}