Educational Codeforces Round 26
Educational Codeforces Round 26
https://codeforces.com/contest/837
D 差一点，补了E
A. Text Volume
#include <bits/stdc++.h>
using namespace std;
int main () {
int n;
string s;
cin >> n;
getchar ();
getline (cin, s);
int ans = 0, len = 0;
for (int i = 0; i < s.size (); i++) {
if (s[i] == ' ') ans = max (ans, len), len = 0;
else if (s[i] >= 'A' && s[i] <= 'Z') len++;
}
ans = max (ans, len);
cout << ans;
}
B. Flag of Berland
#include <bits/stdc++.h>
using namespace std;
const int N = 105;
char a[N][N], idx[] = {'R', 'B', 'G'};
int n, m;
bool check (int model) {
map<char, vector<int>> mp;
if (model) { //横
for (int i = 1; i <= n; i++) {
for (int j = 2; j <= m; j++) {
if (a[i][j] != a[i][j-1]) return false;
}
mp[a[i][1]].push_back (i);
}
}
else {
for (int j = 1; j <= m; j++) {
for (int i = 2; i <= n; i++) {
if (a[i][j] != a[i-1][j]) return false;
}
mp[a[1][j]].push_back (j);
}
}
set<int> s;
for (int i = 0; i < 3; i++) s.insert (mp[idx[i]].size ());
if (s.size () != 1) return false;
//检查是否连续
int nn = *s.begin ();
for (int i = 0; i < 3; i++) {
for (int j = 1; j < nn; j++) {
if (mp[idx[i]][j] - mp[idx[i]][j-1] != 1) return false;
}
}
return true;
}
int main () {
cin >> n >> m;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
cin >> a[i][j];
}
}
if (check (0) || check (1)) cout << "YES\n";
else cout << "NO\n";
}
C. Two Seals
#include <bits/stdc++.h>
using namespace std;
const int N = 105;
int x[N], y[N], a, b, n, ans;
int main () {
cin >> n >> a >> b;
for (int i = 1; i <= n; i++) cin >> x[i] >> y[i];
for (int i = 1; i <= n; i++) {
for (int j = i + 1; j <= n; j++) {
//if (x[i] > a || x[j] > a || y[i] > b || y[j] > b) break;
int s = x[i] * y[i] + x[j] * y[j];
if (x[i] + x[j] <= a && max (y[i], y[j]) <= b) ans = max (ans, s);
if (x[i] + x[j] <= b && max (y[i], y[j]) <= a) ans = max (ans, s);
if (y[i] + y[j] <= a && max (x[i], x[j]) <= b) ans = max (ans, s);
if (y[i] + y[j] <= b && max (x[i], x[j]) <= a) ans = max (ans, s);
if (x[i] + y[j] <= a && max (y[i], x[j]) <= b) ans = max (ans, s);
if (x[i] + y[j] <= b && max (y[i], x[j]) <= a) ans = max (ans, s);
if (y[i] + x[j] <= a && max (x[i], y[j]) <= b) ans = max (ans, s);
if (y[i] + x[j] <= b && max (x[i], y[j]) <= a) ans = max (ans, s);
}
}
cout << ans << endl;
}
//n^2暴力选即可
D. Round Subset
线性dp小题。把一个维度塞进数组，另一个作为值
// LUOGU_RID: 103242598
#include <bits/stdc++.h>
#define ll long long
#define vi vector<ll>
using namespace std;
const int N = 205, M = 30 * N;
ll cnt2[N], cnt5[N];
int n, k;
ll f[N][M][2]; //f[j][k]: 选了j个数,其中有k个5时, 2的数量
int main () {
cin >> n >> k;
for (int i = 1; i <= n; i++) {
ll x;
cin >> x;
while (x % 2 == 0) x /= 2, cnt2[i] ++;
while (x % 5 == 0) x /= 5, cnt5[i] ++;
//cout << cnt2[i] << ' ' << cnt5[i] << endl;
}
//cout << log (1e18) / log (5) << ' ' << log (1e18) / log(2) << endl;
memset (f, -1, sizeof f);
f[0][0][0] = f[0][0][1] = 0;
for (int i = 1; i <= n; i++) {
int t = (i & 1);
for (int j = 0; j <= min(k, i); j++) {
for (int k = 0; k < j * 30; k++) {
if (!j || k < cnt5[i] || f[j-1][k-cnt5[i]][t] == -1) continue;
f[j][k][t^1] = max (f[j][k][t^1], f[j-1][k-cnt5[i]][t] + cnt2[i]);
}
}
//f.swap (tmp);
for (int j = 0; j <= min(k, i); j++) {
for (int k = 0; k < j * 30; k++) {
f[j][k][t] = f[j][k][t^1];
}
}
}
ll ans = 0;
for (ll i = 0; i < 30 * n; i++) { //cnt5
//cout << i << ' ' << f[n][k][i] << endl;
ans = max (ans, min (i, f[k][i][n&1]));
}
cout << ans << endl;
}
//min(cnt2, cnt5)
E. Vasya's Function
#include <bits/stdc++.h>
#define int long long
using namespace std;
typedef pair<int, int> pii; //因子,次数
vector <pii> v;
signed main () {
int x, y, ans = 0;
cin >> x >> y;
int a = x;
//v.push_back ({0, 0});
for (int i = 2; i * i <= x; i++) {
if (x % i == 0) {
v.push_back ({i, 0});
while (x % i == 0) x /= i, v.back ().second ++;
}
}
if (x) v.push_back ({x, 1});
//for (auto i : v) cout << i.first << ' ' << i.second << endl;
x = a;
while (1) {
int maxn = 0, id = 0;
for (int i = 0; i < v.size (); i++) {
auto &p = v[i].first, &cnt = v[i].second;
if (!cnt) continue; //key!!!因为多次迭代会使得因子被用完
int tt = y / p * p; //<y的p的最大倍数
if (tt > maxn) maxn = tt, id = i; //记录最大倍数
}
auto &p = v[id].first, &cnt = v[id].second;
//cout << y << ' ' << maxn << endl;
ans += y - maxn, y = maxn;
if (y == 0) break;
cnt--, x /= p, y /= p;
}
cout << ans << endl;
}
//因子,数量
