codeforces 1096 題解
阿新 • • 發佈：2018-12-29
A：
發現最優的方案一定是選 $ l $ 和 $ 2 * l $，題目保證有解，直接輸出即可
#include <bits/stdc++.h> #define Fast_cin ios::sync_with_stdio(false), cin.tie(); #define rep(i, a, b) for(register int i = a; i <= b; i++) #define per(i, a, b) for(register int i = a; i >= b; i--) #define DEBUG(x) cerr << "DEBUG" << x << " >>> " << endl; using namespace std; typedef unsigned long long ull; typedef long long ll; template <typename _T> inline void read(_T &f) { f = 0; _T fu = 1; char c = getchar(); while(c < '0' || c > '9') { if(c == '-') fu = -1; c = getchar(); } while(c >= '0' && c <= '9') { f = (f << 3) + (f << 1) + (c & 15); c = getchar(); } f *= fu; } template <typename T> void print(T x) { if(x < 0) putchar('-'), x = -x; if(x < 10) putchar(x + 48); else print(x / 10), putchar(x % 10 + 48); } template <typename T> void print(T x, char t) { print(x); putchar(t); } int T, l, r; int main() { read(T); while(T--) { read(l); read(r); print(l, ' '); print(2 * l, '\n'); } return 0; }
B：
情況 1：所有字母都相同，輸出 $ n * (n - 1) / 2 $ 即可
情況 2：左邊有連續 $ x $ 個字母相同，右邊有 $ y $ 個，第一個字母和最後一個字元相同，輸出 $ (x + 1) * (y + 1) $
情況 3：左邊有連續 $ x $ 個字母相同，右邊有 $ y $ 個，第一個字母和最後一個字元不同，輸出 $ x + y + 2 - 1 $，最後的 $ -1 $ 是因為整個串被算了 $ 2 $ 次
#include <bits/stdc++.h> #define Fast_cin ios::sync_with_stdio(false), cin.tie(); #define rep(i, a, b) for(register int i = a; i <= b; i++) #define per(i, a, b) for(register int i = a; i >= b; i--) #define DEBUG(x) cerr << "DEBUG" << x << " >>> " << endl; using namespace std; typedef unsigned long long ull; typedef long long ll; template <typename _T> inline void read(_T &f) { f = 0; _T fu = 1; char c = getchar(); while(c < '0' || c > '9') { if(c == '-') fu = -1; c = getchar(); } while(c >= '0' && c <= '9') { f = (f << 3) + (f << 1) + (c & 15); c = getchar(); } f *= fu; } template <typename T> void print(T x) { if(x < 0) putchar('-'), x = -x; if(x < 10) putchar(x + 48); else print(x / 10), putchar(x % 10 + 48); } template <typename T> void print(T x, char t) { print(x); putchar(t); } const int N = 2e5 + 5, md = 998244353; char c[N]; int cnt[23333]; int n, cut1 = -1, cut2 = -1; inline int mul(int x, int y) { return (ll)x * y % md; } int main() { read(n); scanf("%s", c + 1); for(register int i = 2; i <= n; i++) if(c[i] != c[i - 1]) { cut1 = i - 1; break; } for(register int i = n - 1; i >= 1; i--) if(c[i] != c[i + 1]) { cut2 = i + 1; break; } if(cut1 == -1) cout << (ll)n * (n - 1) / 2 % md << endl; else if(c[1] == c[n]) cout << mul(cut1 + 1, n - cut2 + 2) << endl; else cout << cut1 + 1 + n - cut2 + 2 - 1 << endl; return 0; }
C：
答案本質上是把一個圓切成答案份
那麼圓心角確定了，就可以算出一個圓周角的大小，如果切成 360 份可以拼出任意角度，所以列舉這個角度即可
#include <bits/stdc++.h> #define Fast_cin ios::sync_with_stdio(false), cin.tie(); #define rep(i, a, b) for(register int i = a; i <= b; i++) #define per(i, a, b) for(register int i = a; i >= b; i--) #define DEBUG(x) cerr << "DEBUG" << x << " >>> " << endl; using namespace std; typedef unsigned long long ull; typedef long long ll; template <typename _T> inline void read(_T &f) { f = 0; _T fu = 1; char c = getchar(); while(c < '0' || c > '9') { if(c == '-') fu = -1; c = getchar(); } while(c >= '0' && c <= '9') { f = (f << 3) + (f << 1) + (c & 15); c = getchar(); } f *= fu; } template <typename T> void print(T x) { if(x < 0) putchar('-'), x = -x; if(x < 10) putchar(x + 48); else print(x / 10), putchar(x % 10 + 48); } template <typename T> void print(T x, char t) { print(x); putchar(t); } int T; int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); } int main() { read(T); while(T--) { int d; read(d); int ans = -1; for(register int i = 3; i <= 23333; i++) { if(d * i % 180 == 0 && d <= 180 - 360 / (double)i) { ans = i; break; } } print(ans, '\n'); } return 0; }
D：
$ f[i][j] $ 表示到了第 $ i $ 個字母，$ hard $ 已經匹配到了第 $ j $ 個字母的最小代價
直接 $ dp $ 即可
#include <bits/stdc++.h>
#define int long long
#define Fast_cin ios::sync_with_stdio(false), cin.tie();
#define rep(i, a, b) for(register int i = a; i <= b; i++)
#define per(i, a, b) for(register int i = a; i >= b; i--)
#define DEBUG(x) cerr << "DEBUG" << x << " >>> " << endl;
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
template <typename _T>
inline void read(_T &f) {
f = 0; _T fu = 1; char c = getchar();
while(c < '0' || c > '9') { if(c == '-') fu = -1; c = getchar(); }
while(c >= '0' && c <= '9') { f = (f << 3) + (f << 1) + (c & 15); c = getchar(); }
f *= fu;
}
template <typename T>
void print(T x) {
if(x < 0) putchar('-'), x = -x;
if(x < 10) putchar(x + 48);
else print(x / 10), putchar(x % 10 + 48);
}
template <typename T>
void print(T x, char t) {
print(x); putchar(t);
}
const int N = 1e5 + 5;
int f[N][5], w[N];
char c[N];
int n;
int calc(char t) {
if(t == 'h') return 1;
if(t == 'a') return 2;
if(t == 'r') return 3;
if(t == 'd') return 4;
return 0;
}
#undef int
int main() {
#define int long long
memset(f, -1, sizeof(f));
read(n); scanf("%s", c + 1);
for(register int i = 1; i <= n; i++) read(w[i]);
f[0][0] = 0;
for(register int i = 0; i < n; i++) {
int val = calc(c[i + 1]);
for(register int j = 0; j <= 3; j++) {
if(f[i][j] == -1) continue;
if(val == j + 1) {
if(f[i + 1][j + 1] == -1) f[i + 1][j + 1] = f[i][j];
else f[i + 1][j + 1] = min(f[i + 1][j + 1], f[i][j]);
if(f[i + 1][j] == -1) f[i + 1][j] = f[i][j] + w[i + 1];
else f[i + 1][j] = min(f[i + 1][j], f[i][j] + w[i + 1]);
} else {
if(f[i + 1][j] == -1) f[i + 1][j] = f[i][j];
else f[i + 1][j] = min(f[i + 1][j], f[i][j]);
}
}
}
int ans = f[n][0];
for(register int i = 0; i <= 3; i++) if(f[n][i] != -1) ans = min(ans, f[n][i]);
cout << ans << endl;
return 0;
}E：
沒寫出來，先咕了
F：
分別計算 $ -1 $ 和 $ -1 $ 的貢獻，$ -1 $ 和數字的貢獻，數字和數字的貢獻
第一個用 $ dp $ 求出，第二個用字首和求出，第三個用樹狀陣列求出
$ sb $ 的我能用字首和的地方寫了樹狀陣列
#include <bits/stdc++.h>
#define Fast_cin ios::sync_with_stdio(false), cin.tie();
#define rep(i, a, b) for(register int i = a; i <= b; i++)
#define per(i, a, b) for(register int i = a; i >= b; i--)
#define DEBUG(x) cerr << "DEBUG" << x << " >>> " << endl;
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
template <typename _T>
inline void read(_T &f) {
f = 0; _T fu = 1; char c = getchar();
while(c < '0' || c > '9') { if(c == '-') fu = -1; c = getchar(); }
while(c >= '0' && c <= '9') { f = (f << 3) + (f << 1) + (c & 15); c = getchar(); }
f *= fu;
}
template <typename T>
void print(T x) {
if(x < 0) putchar('-'), x = -x;
if(x < 10) putchar(x + 48);
else print(x / 10), putchar(x % 10 + 48);
}
template <typename T>
void print(T x, char t) {
print(x); putchar(t);
}
const int N = 2e5 + 5, md = 998244353;
inline int mul(int x, int y) { return (ll)x * y % md; }
inline int add(int x, int y) {
x += y;
if(x >= md) x -= md;
return x;
}
inline int sqr(int x) { return mul(x, x); }
inline int fpow(int x, int y) {
int ans = 1;
while(y) {
if(y & 1) ans = mul(ans, x);
y >>= 1; x = sqr(x);
}
return ans;
}
int a[N], s[N], f[N], fac[N];
int n, inv2 = (md + 1) / 2, ans1;
inline int lowbit(int x) { return x & -x; }
void change(int x, int y) {
for(register int i = x; i <= n; i += lowbit(i))
f[i] += y;
}
int query(int x) {
int ans = 0;
for(register int i = x; i; i -= lowbit(i))
ans += f[i];
return ans;
}
int main() {
read(n); fac[0] = 1;
for(register int i = 1; i <= n; i++) read(a[i]), fac[i] = mul(fac[i - 1], i), s[i] = 1;
for(register int i = n; i >= 1; i--) {
if(a[i] == -1) continue;
ans1 = add(ans1, query(a[i]));
change(a[i], 1);
}
memset(f, 0, sizeof(f));
for(register int i = 1; i <= n; i++) {
if(a[i] != -1) s[a[i]] = 0;
}
for(register int i = 1; i <= n; i++) s[i] += s[i - 1];
ans1 = mul(ans1, fac[s[n]]);
for(register int i = 1; i <= s[n]; i++) f[i] = add(f[i - 1], mul(mul(fac[s[n]], fpow(i, md - 2)), mul(i - 1, mul(i, inv2))));
ans1 = add(ans1, f[s[n]]); memset(f, 0, sizeof(f));
for(register int i = 1; i <= n; i++) if(a[i] == -1) change(i, 1);
for(register int i = 1; i <= n; i++) {
if(a[i] == -1) continue;
int val = mul(s[a[i]], query(n) - query(i));
val = add(val, mul(s[n] - s[a[i]], query(i)));
val = mul(val, fac[s[n] - 1]);
ans1 = add(ans1, val);
}
print(mul(ans1, fpow(fac[s[n]], md - 2)), '\n');
return 0;
}G：
看到揹包問題就會想到卷積，因為揹包的轉移和卷積的形式相同
對於最開始的 $ k $ 個數構造生成函式，計算 $ n / 2 $ 次冪，每一位的平方加起來就是答案啦
#pragma GCC target("avx")
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
#define Fast_cin ios::sync_with_stdio(false), cin.tie();
#define rep(i, a, b) for(register int i = a; i <= b; i++)
#define per(i, a, b) for(register int i = a; i >= b; i--)
#define DEBUG(x) cerr << "DEBUG" << x << " >>> " << endl;
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
template <typename _T>
inline void read(_T &f) {
f = 0; _T fu = 1; char c = getchar();
while(c < '0' || c > '9') { if(c == '-') fu = -1; c = getchar(); }
while(c >= '0' && c <= '9') { f = (f << 3) + (f << 1) + (c & 15); c = getchar(); }
f *= fu;
}
template <typename T>
void print(T x) {
if(x < 0) putchar('-'), x = -x;
if(x < 10) putchar(x + 48);
else print(x / 10), putchar(x % 10 + 48);
}
template <typename T>
void print(T x, char t) {
print(x); putchar(t);
}
const int P = 998244353;
inline int add(int x, int y) {
x += y;
if(x >= P) x -= P;
return x;
}
inline int sub(int x, int y) {
x -= y;
if(x < 0) x += P;
return x;
}
inline int mul(int x, int y) {
return (ll)x * y % P;
}
inline int fpow(int x, int y) {
int ans = 1;
while(y) {
if(y & 1) ans = mul(ans, x);
y >>= 1; x = mul(x, x);
}
return ans;
}
namespace ntt {
int base = 1, root = -1, maxbase = -1;
vector <int> roots = {0, 1}, rev = {0, 1};
void init() {
int tmp = P - 1; maxbase = 0;
while(!(tmp & 1)) {
tmp >>= 1;
maxbase++;
}
root = 2;
while(1) {
if(fpow(root, 1 << maxbase) == 1 && fpow(root, 1 << (maxbase - 1)) != 1) break;
root++;
}
}
void ensure_base(int nbase) {
if(maxbase == -1) init();
if(nbase <= base) return;
assert(nbase <= maxbase);
rev.resize(1 << nbase);
for(register int i = 1; i < (1 << nbase); i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (nbase - 1));
roots.resize(1 << nbase);
while(base < nbase) {
int z = fpow(root, 1 << (maxbase - base - 1));
for(register int i = (1 << (base - 1)); i < (1 << base); i++) {
roots[i << 1] = roots[i];
roots[i << 1 | 1] = mul(roots[i], z);
}
base++;
}
}
void dft(vector <int> &a) {
int n = a.size(), zeros = __builtin_ctz(n);
ensure_base(zeros);
int shift = base - zeros;
for(register int i = 0; i < n; i++) if(i < (rev[i] >> shift)) swap(a[i], a[rev[i] >> shift]);
for(register int mid = 1; mid < n; mid <<= 1) {
for(register int i = 0; i < n; i += (mid << 1)) {
for(register int j = 0; j < mid; j++) {
int x = a[i + j], y = mul(a[i + j + mid], roots[mid + j]);
a[i + j] = add(x, y); a[i + j + mid] = sub(x, y);
}
}
}
}
vector <int> pmul(vector <int> a, vector <int> b, bool is_sqr = false) {
int need = a.size() + b.size() - 1, nbase = 0;
while((1 << nbase) < need) nbase++;
ensure_base(nbase); int size = 1 << nbase;
a.resize(size); dft(a); if(!is_sqr) b.resize(size), dft(b); else b = a;
int inv = fpow(size, P - 2);
for(register int i = 0; i < size; i++) a[i] = mul(a[i], mul(b[i], inv));
reverse(a.begin() + 1, a.end()); dft(a); a.resize(need); return a;
}
vector <int> psqr(vector <int> a) { return pmul(a, a, 1); }
vector <int> inv(vector <int> a, int size) {
if(size == 1) return vector <int> { fpow(a[0], P - 2) };
vector <int> b = inv(a, size >> 1); a = pmul(a, psqr(b)); b.resize(size);
for(register int i = 0; i < size; i++) b[i] = sub(add(b[i], b[i]), a[i]);
return b;
}
vector <int> pinv(vector <int> a) {
int nbase = 0; while((1 << nbase) < a.size()) nbase++;
return inv(a, 1 << nbase);
}
}
using ntt::pmul;
using ntt::psqr;
vector <int> wxw, ans;
int n, k, sum;
int main() {
read(n); read(k); wxw.resize(10);
for(register int i = 1; i <= k; i++) {
int t; read(t); wxw[t] = 1;
}
n >>= 1; ans = wxw; n--;
while(n) {
if(n & 1) ans = pmul(ans, wxw);
n >>= 1; wxw = psqr(wxw);
}
for(register int i = 0; i < ans.size(); i++) sum = add(sum, mul(ans[i], ans[i]));
cout << sum << endl;
return 0;
}