Codeforces Round #558 (Div. 2)
目录
Codeforces Round #558 (Div. 2) 题解
A Eating Soup
分类讨论一下
\(m\) 个断点最多可以将环分成 \(m\) 段，而剩下 \(n-m\) 个点最多可以成为 \(n-m\) 段，取个 \(\min\) 即可
注意 \(m=0\) 时是一个环答案为 \(1\)
// Copyright lzt
#include<stdio.h>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<cmath>
#include<iostream>
#include<queue>
#include<string>
#include<ctime>
using namespace std;
typedef long long ll;
typedef std::pair<int, int> pii;
typedef long double ld;
typedef unsigned long long ull;
typedef std::pair<long long, long long> pll;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define rep(i, j, k) for (register int i = (int)(j); i <= (int)(k); i++)
#define rrep(i, j, k) for (register int i = (int)(j); i >= (int)(k); i--)
#define Debug(...) fprintf(stderr, __VA_ARGS__)
inline ll read() {
ll x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch <= '9' && ch >= '0') {
x = 10 * x + ch - '0';
ch = getchar();
}
return x * f;
}
int n, m;
void work() {
n = read(), m = read();
if (m == 0) {
puts("1");
return;
}
int num = n - m;
printf("%d\n", min(num, m));
}
int main() {
#ifdef LZT
freopen("in", "r", stdin);
#endif
work();
#ifdef LZT
Debug("My Time: %.3lfms\n", (double)clock() / CLOCKS_PER_SEC);
#endif
}
B Cat Party
倒着枚举长度并且维护每一个出现次数的个数
就是维护出现次数为 \(i\) 的数字有多少个
显然出现次数最多只能有两种（因为你只能修改 \(1\) 个数，也就只能修改一个出现次数）
然后就分类讨论一下最后删掉的是哪个即可
// Copyright lzt
#include<stdio.h>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<cmath>
#include<iostream>
#include<queue>
#include<string>
#include<ctime>
using namespace std;
typedef long long ll;
typedef std::pair<int, int> pii;
typedef long double ld;
typedef unsigned long long ull;
typedef std::pair<long long, long long> pll;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define rep(i, j, k) for (register int i = (int)(j); i <= (int)(k); i++)
#define rrep(i, j, k) for (register int i = (int)(j); i >= (int)(k); i--)
#define Debug(...) fprintf(stderr, __VA_ARGS__)
inline ll read() {
ll x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch <= '9' && ch >= '0') {
x = 10 * x + ch - '0';
ch = getchar();
}
return x * f;
}
const int maxn = 100100;
int n, cnt;
int a[maxn];
int b[maxn], num[maxn];
void work() {
n = read();
rep(i, 1, n) a[i] = read(), b[a[i]]++;
cnt = 0; set<int> col;
rep(i, 1, 100000) {
if (b[i]) {
num[b[i]]++;
cnt++, col.insert(b[i]);
}
}
rrep(i, n, 2) {
int nw = *col.begin();
if (num[nw] == 1 && col.size() > 1) nw = *(++col.begin());
if (col.size() <= 2) {
bool ok = 0;
if (col.size() == 1) {
int nw = *col.begin();
if (nw == 1 || num[nw] == 1) ok = 1;
else ok = 0;
} else {
int A = *col.begin(), B = *(++col.begin());
if (num[A] == 1 && A == 1) ok = 1;
else if (num[B] == 1 && B == 1) ok = 1;
else if (num[B] == 1 && B == A + 1) ok = 1;
else ok = 0;
}
if (ok) {
printf("%d\n", i);
return;
}
}
num[b[a[i]]]--;
if (num[b[a[i]]] == 0) col.erase(b[a[i]]);
b[a[i]]--;
if (b[a[i]]) {
num[b[a[i]]]++;
if (num[b[a[i]]] == 1) col.insert(b[a[i]]);
}
else cnt--;
}
puts("1");
}
int main() {
#ifdef LZT
freopen("in", "r", stdin);
#endif
work();
#ifdef LZT
Debug("My Time: %.3lfms\n", (double)clock() / CLOCKS_PER_SEC);
#endif
}
C Power Transmission
答案显然是 \(calc(\text{不同的直线数量}) - \sum_{\text{斜率k}} calc(\text{斜率为k的直线数量})\)
其中 \(calc(x)=x*(x-1)/2\)
然后就枚举每两个点形成一条线，然后计数即可
// Copyright lzt
#include<stdio.h>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<cmath>
#include<iostream>
#include<queue>
#include<string>
#include<ctime>
using namespace std;
typedef long long ll;
typedef std::pair<int, int> pii;
typedef long double ld;
typedef unsigned long long ull;
typedef std::pair<long long, long long> pll;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define rep(i, j, k) for (register int i = (int)(j); i <= (int)(k); i++)
#define rrep(i, j, k) for (register int i = (int)(j); i >= (int)(k); i--)
#define Debug(...) fprintf(stderr, __VA_ARGS__)
inline ll read() {
ll x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch <= '9' && ch >= '0') {
x = 10 * x + ch - '0';
ch = getchar();
}
return x * f;
}
const int maxn = 1010;
int n;
pii p[maxn];
bool sm(double x, double y) {
return fabs(x - y) <= 1e-8;
}
struct Line {
// Ax + By + C = 0
double A, B, C;
bool operator == (const Line &b) const {
return sm(A, b.A) && sm(B, b.B) && sm(C, b.C);
}
bool operator < (const Line &b) const {
if (sm(A, b.A)) {
if (sm(B, b.B)) return C < b.C;
return B < b.B;
}
return A < b.A;
}
void init() {
if (A == 0) C = C / B, B = 1;
else B = B / A, C = C / A, A = 1;
}
double calc() {
init();
if (A == 1) return B;
else return 1e30;
}
};
map<double, map<double, int> > M;
map<double, int> shit;
vector<Line> v;
void work() {
n = read();
rep(i, 1, n) p[i].fi = read(), p[i].se = read();
ll num = 0;
rep(i, 1, n) rep(j, i + 1, n) {
Line nw;
nw.A = p[j].se - p[i].se;
nw.B = p[i].fi - p[j].fi;
nw.C = 0 - p[i].fi * nw.A - p[i].se * nw.B;
nw.init();
double k = nw.calc();
if (M[k][nw.C] == 0) {
M[k][nw.C] = 1;
num++; shit[k]++;
}
}
num = num * (num - 1) / 2;
for (map<double, int>::iterator it = shit.begin(); it != shit.end(); it++) {
int nw = it -> se;
num = num - nw * 1ll * (nw - 1) / 2;
}
printf("%lld\n", num);
}
int main() {
#ifdef LZT
freopen("in", "r", stdin);
#endif
work();
#ifdef LZT
Debug("My Time: %.3lfms\n", (double)clock() / CLOCKS_PER_SEC);
#endif
}
D Mysterious Code
首先预处理 \(S[i][j]\) 和 \(T[i][j]\) 分别表示当前串的后缀与 \(s/t\) 的前缀的最长公共子串长度为 \(i\) （也就是只有最后 \(i\) 位有用，前面的都忽略），然后下一个位置填了字母 \(j\) 之后那个最长公共子串的长度（类似 \(kmp\) 的 \(fail\)）
直接 \(dp\) \(f[i][j][k]\) 表示当前长度为 \(i\)，在第一个串上的位置是 \(j\) ，在第二个串上的位置是 \(k\) 的最大 \(f(c',s)-f(c',t)\)
然后转移就暴力枚举一下下一个是啥
// Copyright lzt
#include<stdio.h>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<cmath>
#include<iostream>
#include<queue>
#include<string>
#include<ctime>
using namespace std;
typedef long long ll;
typedef std::pair<int, int> pii;
typedef long double ld;
typedef unsigned long long ull;
typedef std::pair<long long, long long> pll;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define rep(i, j, k) for (register int i = (int)(j); i <= (int)(k); i++)
#define rrep(i, j, k) for (register int i = (int)(j); i >= (int)(k); i--)
#define Debug(...) fprintf(stderr, __VA_ARGS__)
inline ll read() {
ll x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch <= '9' && ch >= '0') {
x = 10 * x + ch - '0';
ch = getchar();
}
return x * f;
}
const int maxn = 1010;
int n, m, len;
char s[maxn], t[maxn], c[maxn];
int f[maxn][55][55];
int S[55][26], T[55][26];
inline void getmx(int &x, int y) {
if (x < y) x = y;
}
void work() {
scanf("%s", c + 1); len = strlen(c + 1);
scanf("%s", s + 1); n = strlen(s + 1);
rep(i, 0, n) rep(j, 0, 25) {
int len = i + 1;
rrep(k, min(len, n), 1) {
bool flag = true;
rep(_, 1, k - 1) if (s[i - k + _ + 1] != s[_]) {
flag = false; break;
}
if (s[k] - 'a' != j) flag = false;
if (flag) {
S[i][j] = k;
break;
}
}
// cout<<i<<' '<<j<<' '<<S[i][j]<<endl;
}
scanf("%s", t + 1); m = strlen(t + 1);
rep(i, 0, m) rep(j, 0, 25) {
int len = i + 1;
rrep(k, min(len, m), 1) {
bool flag = true;
rep(_, 1, k - 1) if (t[i - k + _ + 1] != t[_]) {
flag = false; break;
}
if (t[k] - 'a' != j) flag = false;
if (flag) {
T[i][j] = k;
break;
}
}
// cout<<i<<' '<<j<<' '<<T[i][j]<<endl;
}
rep(i, 0, len) rep(j, 0, n) rep(k, 0, m) f[i][j][k] = -1e9;
f[0][0][0] = 0;
rep(i, 0, len - 1) rep(j, 0, n) rep(k, 0, m) {
if (f[i][j][k] == -1e9) continue;
if (c[i + 1] != '*') {
int nw = c[i + 1] - 'a', pl = 0;
if (S[j][nw] == n) pl++;
if (T[k][nw] == m) pl--;
getmx(f[i + 1][S[j][nw]][T[k][nw]], f[i][j][k] + pl);
} else {
rep(nw, 0, 25) {
int pl = 0;
if (S[j][nw] == n) pl++;
if (T[k][nw] == m) pl--;
getmx(f[i + 1][S[j][nw]][T[k][nw]], f[i][j][k] + pl);
}
}
}
int ans = -1e9;
rep(j, 0, n) rep(k, 0, m) {
getmx(ans, f[len][j][k]);
}
printf("%d\n", ans);
}
int main() {
#ifdef LZT
freopen("in", "r", stdin);
#endif
work();
#ifdef LZT
Debug("My Time: %.3lfms\n", (double)clock() / CLOCKS_PER_SEC);
#endif
}
E Magical Permutation
// Copyright lzt
#include<stdio.h>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<cmath>
#include<iostream>
#include<queue>
#include<string>
#include<ctime>
using namespace std;
typedef long long ll;
typedef std::pair<int, int> pii;
typedef long double ld;
typedef unsigned long long ull;
typedef std::pair<long long, long long> pll;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define rep(i, j, k) for (register int i = (int)(j); i <= (int)(k); i++)
#define rrep(i, j, k) for (register int i = (int)(j); i >= (int)(k); i--)
#define Debug(...) fprintf(stderr, __VA_ARGS__)
inline ll read() {
ll x = 0, f = 1;
char ch = getchar();
while (ch < '0' || ch > '9') {
if (ch == '-') f = -1;
ch = getchar();
}
while (ch <= '9' && ch >= '0') {
x = 10 * x + ch - '0';
ch = getchar();
}
return x * f;
}
const int maxn = 200200;
int n, ans, pos;
int a[maxn], lb[20];
vector<int> vec;
bool vis[maxn << 2];
void add(int v) {
int tmp = v;
rrep(i, 20, 0) {
if (v & (1 << i)) {
if (lb[i]) v ^= lb[i];
else {
vec.pb(tmp);
lb[i] = v;
break;
}
}
}
}
void dfs(int v, int num) {
printf("%d ", v); vis[v] = 1;
if (num == (1 << ans)) return;
rep(i, 0, vec.size() - 1) if (!vis[v ^ vec[i]]) {
dfs(v ^ vec[i], num + 1);
break;
}
}
void work() {
n = read();
rep(i, 1, n) a[i] = read();
sort(a + 1, a + n + 1);
ans = 0, pos = 1;
rep(i, 1, 20) {
while (pos <= n && a[pos] < (1 << i)) {
add(a[pos]);
pos++;
}
bool flag = true;
rep(j, 0, i - 1) if (!lb[j]) flag = false;
if (flag) ans = i;
}
memset(lb, 0, sizeof(lb));
vec.clear();
rep(i, 1, n) {
if (a[i] < (1 << ans)) add(a[i]);
}
printf("%d\n", ans);
dfs(0, 1);
}
int main() {
#ifdef LZT
freopen("in", "r", stdin);
#endif
work();
#ifdef LZT
Debug("My Time: %.3lfms\n", (double)clock() / CLOCKS_PER_SEC);
#endif
}
