2020杭电多校第一场
1004.Distinct Sub-palindromes
签到题，问长度为n且回文子串的个数最少的串的数量
想到 abcabcabc... 排列回文子串最少为3
n=1，2，3特殊处理一下
#include <bits/stdc++.h> using namespace std; typedef long long ll; #define rep(i, a, b) for (register int i = a; i <= b; i++) ll n; void solve() { cin >> n; if (n == 1) puts("26"); else if (n == 2) puts("676"); else if (n == 3) cout << 26 * 26 * 26 << endl; else cout << 26 * 25 * 24 << endl; } int main() { int t = 1; cin >> t; while (t--) { solve(); } }
1005. Fibonacci Sum
题意：求这个式子
先化简这个式子，根据斐波拉契通项：
化简完毕，求x2 ≡ 5（mod 1e9+9）得 x=383008016（二次剩余）
所以可求得1/√5 ≡ 276601605（mod 1e9+9），a ≡ 691504013（mod 1e9+9）， b ≡ 308495997（mod 1e9+9）
求A，B，qn的时候用欧拉降幂即可，还有亿点点小细节看代码。(代码t了，大佬能指点一下吗)
#include <bits/stdc++.h> using namespace std; typedef long long ll; #define rep(i, a, b) for (register int i = a; i <= b; i++) int mod = 1e9 + 9; inline int ksm(int a, int b) { int res = 1; for (; b; b >>= 1, a = (ll)a * a % mod) if (b & 1) res = (ll)res * a % mod; return res; } int fac[100010], inv[100010]; inline int C(int k, int j) { return (ll)fac[k] * inv[j] % mod * inv[k - j] % mod; } int x = 383008016, x2 = 276601605, a = 691504013, b = 308495997; int A, B, q, Q, d, D, tmp; //Q = q ^ (n + 1), d = B / A, D = d ^ (n + 1) ll n, c, k; int ans; inline void solve() { cin >> n >> c >> k; ans = 0; A = ksm(a, c % (mod - 1)); B = ksm(b, c % (mod - 1)); q = ksm(A, k); Q = ksm(q, (n + 1) % (mod - 1)); d = (ll)ksm(A, mod - 2) * B % mod; D = ksm(d, (n + 1) % (mod - 1)); rep(j, 0, k) { tmp = q == 1 ? n % mod : (ll)(Q - q + mod) % mod * ksm(q - 1, mod - 2) % mod; tmp = (ll)tmp * C(k, j) % mod; tmp *= j & 1 ? -1 : 1; ans = (ans + tmp + mod) % mod; q = (ll)q * d % mod; Q = (ll)Q * D % mod; } cout << (ll)ksm(x2, k) * ans % mod << endl; } int main() { ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0); fac[0] = inv[0] = 1; rep(i, 1, 100000) fac[i] = (ll)fac[i - 1] * i % mod; inv[100000] = ksm(fac[100000], mod - 2); for (int i = 99999; i > 0; i--) inv[i] = (ll)inv[i + 1] * (i + 1) % mod; int t = 1; cin >> t; while (t--) { solve(); } }
1009.Leading Robots
要判断哪些robot能领先。
先对加速度从大到小排序，再舍掉初始位置比之前小的（加速度小，初始位置小不可能追上）
这样后面的bot一定能追上前面的bot，然后如果当前bot在追上下一个bot之前就被追上了，那么他也要被舍弃
最后再排除掉初始位置和加速度相同的bot
#include <bits/stdc++.h> using namespace std; typedef long long ll; #define rep(i, a, b) for (register int i = a; i <= b; i++) #define pii pair<int, int> struct rob { ll a, p; bool operator<(rob &f) const { return a == f.a ? p > f.p : a > f.a; } } c[50010], tmp[50010], frot[50010]; double tim(rob x, rob y) { return 1.0 * (x.p - y.p) / (x.a - y.a); } int n; map<pair<int, int>, int> mp; ll s, cnt, id, ans; void solve() { mp.clear(); cin >> n; rep(i, 1, n) { cin >> c[i].a >> c[i].p; mp[make_pair(c[i].a, c[i].p)]++; } sort(c + 1, c + n + 1); s = cnt = id = ans = 0; rep(i, 1, n) if (c[i].p > s) { s = c[i].p; tmp[++cnt] = c[i]; } rep(i, 1, cnt) { while (id > 1 && tim(tmp[i], frot[id]) <= tim(frot[id], frot[id - 1])) id--; frot[++id] = tmp[i]; } rep(i, 1, id) if (mp[make_pair(frot[i].a, frot[i].p)] == 1) ans++; cout << ans << endl; } int main() { ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0); int t = 1; cin >> t; while (t--) { solve(); } }
1011.Minimum Index
求字符串每个前缀的最小后缀。
涉及到 Lyndon Word
#include <bits/stdc++.h> using namespace std; typedef long long ll; #define rep(i, a, b) for (register int i = a; i <= b; i++) int mod = 1e9 + 7; int n; int ans[1000010]; int sum, tmp; void solve() { string s; cin >> s; n = s.length(); sum = 1; tmp = 1112; int k = 0; while (k < n) { int i, j; ans[k] = k; for (i = k, j = i + 1; j < n && s[j] >= s[i]; j++) if (s[j] > s[i]) ans[j] = i = k; else { ans[j] = ans[i] + j - i; i++; } while (k <= i) k += j - i; } rep(i, 1, n - 1) { sum = (sum + 1ll * (ans[i] + 1) * tmp) % mod; tmp = (tmp * 1112ll) % mod; } cout << sum << endl; } int main() { ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0); int t = 1; cin >> t; while (t--) { solve(); } }
