Codeforces Round #828 (Div. 3)
Codeforces Round #828 (Div. 3)
https://codeforces.com/contest/1744
罚时爆炸的a~e1
A. Number Replacement
数字一样的对应字母一定要一样。
我写的好复杂，不知道为啥脑抽了，写了巨久
#include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> pii;
const int N = 55;
void solve () {
int n;
cin >> n;
vector <int> v[N];
for (int i = 0; i < n; i++) {
int x;
cin >> x;
v[x].push_back (i);
}
string s;
cin >> s;
for (int i = 1; i <= 50; i++) {
if (v[i].empty ()) continue;
for (int j = 1; j < v[i].size (); j++) {
if (s[v[i][j]] != s[v[i][j-1]]) {
puts ("NO");
return ;
}
}
}
puts ("YES");
}
int main () {
int t;
cin >> t;
while (t --) solve ();
}
B. Even-Odd Increments
奇偶数加减性质，详见代码注释
#include <bits/stdc++.h>
#define int long long
using namespace std;
typedef pair<int, int> pii;
const int N = 1e5 + 5;
int a[N];
void solve () {
int n, q, ans = 0, odd = 0, even = 0;
cin >> n >> q;
for (int i = 1; i <= n; i++) {
cin >> a[i], ans += a[i];
if (a[i] & 1) odd ++;
else even ++;
}
while (q --) {
int op, x;
cin >> op >> x;
if (op == 0) {
//偶+x
if (x & 1) {
//偶 + 奇数 = 奇
ans += even * x;
//全变奇
odd = n, even = 0;
}
else {
//偶 + 偶 = 偶
ans += even * x;
}
}
else {
if (x & 1) {
//奇 + 奇 = 偶
ans += odd * x;
//全变偶
even = n, odd = 0;
}
else {
// 奇 + 偶 = 奇
ans += odd * x;
}
}
cout << ans << endl;
}
}
signed main () {
int t;
cin >> t;
while (t --) solve ();
}
C. Traffic Light
二分到离他最近的green
脑抽，T了两发
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int a[N];
void solve () {
int n;
char c;
string s;
cin >> n >> c >> s;
if (c == 'g') {
cout << "0\n";
return ;
}
set <int> ss;
for (int i = 0; i < n; i++) {
if (s[i] == 'g') ss.insert (i);
}
//s += s;
int ans = 0, tt = *ss.begin ();
for (int i = 0; i < n; i++) {
if (s[i] == c) {
auto it = ss.upper_bound (i);
if (it == ss.end ()) ans = max (ans, n - i + tt);
else ans = max (ans, *it - i);
}
}
cout << ans << "\n";
}
signed main () {
ios::sync_with_stdio (0);cin.tie (0);cout.tie (0);
int t;
cin >> t;
while (t --) solve ();
}
D. Divisibility by 2^n
首先分解每个数，看能拼出多少个2，统计因子2的个数为cnt，如果cnt>=n，则不需要额外操作
否则对于2,4,6,8,...偶数位上的数*i能增加对应的2因子数，即对n以内的偶数（2,4,6,...,n或n-1）再进行2的因数分解，按照所含2因数的个数降序排序，然后挨个加，直至cnt>=n即可
#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 2e5 + 5;
void solve () {
int n, cnt = 0; //2的个数, 额外的
cin >> n;
vector <int> v;
for (int i = 1; i <= n; i++) {
int x;
cin >> x;
while (x % 2 == 0) x /= 2, cnt ++;
int dx = 0, j = i;
while (j % 2 == 0) j /= 2, dx ++;
if (dx) v.push_back (dx);
}
if (cnt >= n) {
cout << "0\n";
return ;
}
int res = n - cnt;
sort (v.begin (), v.end (), greater <int> ());
int pos = 0, ans = 0;
while (res > 0 && pos < v.size ()) {
res -= v[pos ++];
ans ++;
}
if (res > 0) cout << "-1\n";
else cout << ans << "\n";
}
signed main () {
ios::sync_with_stdio (0);cin.tie (0);cout.tie (0);
int t;
cin >> t;
while (t --) solve ();
}
//看缺几个2，补就完了
E1. Divisible Numbers (easy version)
枚举x,利用gcd的性质来check y的合法性
#include <bits/stdc++.h>
#define int long long
using namespace std;
void solve () {
int a, b, c, d;
cin >> a >> b >> c >> d;
int m = a * b;
if (c - a <= b - d) {
for (int i = a + 1; i <= c; i++) {
int ans = m / __gcd (m, i);
//cout << "ans=" << ans << endl;
if (ans > d) continue;
int base = ans;
for (int j = 1; ans <= d; j++) {
ans = base * j;
//cout << "ans=" << ans << endl;
if (ans > b) break;
}
if (ans <= d) {
cout << i << ' ' << ans << "\n";
return ;
}
}
cout << "-1 -1\n";
}
else {
for (int i = b + 1; i <= d; i++) {
int ans = m / __gcd (m, i);
//cout << "ans=" << ans << endl;
if (ans > c) continue;
int base = ans;
for (int j = 1; ans <= c; j++) {
ans = base * j;
//cout << "ans=" << ans << endl;
if (ans > a) break;
}
if (ans <= c) {
cout << ans << ' ' << i << "\n";
return ;
}
}
cout << "-1 -1\n";
}
}
signed main () {
ios::sync_with_stdio (0);cin.tie (0);cout.tie (0);
int t;
cin >> t;
while (t --) solve ();
}
//交叉构造
E2. Divisible Numbers (hard version)
因数分解，赛时不知道怎么处理因数的组合。
其实可以用背包思想，map来更新它的合法性
详见代码
#include <bits/stdc++.h>
#define int long long
using namespace std;
void get_div (int x, vector <int> &v) {
for (int i = 2; i * i <= x; i++) {
while (x % i == 0) {
v.push_back (i);
x /= i;
}
}
if (x > 1) v.push_back (x);
}
void solve () {
int a, b, c, d;
cin >> a >> b >> c >> d;
int m = a * b;
vector <int> v;
get_div (a, v), get_div (b, v);
map <int, bool> dp; //储存c范围内的素因子组合
dp[1] = true;
for (auto i : v) { //枚举a*b的所有因子
auto tmp = dp;
for (auto j : dp) {
if (i * j.first <= c) tmp[i * j.first] = true;
}
dp = tmp;
}
//for (auto i : dp) cout << i.first << ' ';cout << endl;
for (auto i : dp) {
int x = i.first;
int y = m / x;
x = x * (c / x), y = y * (d / y); //限定范围下的上界
if (x > a && y > b) {
if (x <= c && y <= d && (x * y) % (a * b) == 0) {
cout << x << ' ' << y << "\n";
return ;
}
}
}
cout << "-1 -1\n";
}
signed main () {
ios::sync_with_stdio (0);cin.tie (0);cout.tie (0);
int t;
cin >> t;
while (t --) solve ();
}
//质因数分解
F. MEX vs MED
重要性质：要满足 mex > med，必须有区间长度 <= 2 * mex
简单说明：mex = i 时，代表了区间l,r出现了0~i-1的数，那么这些数总共就占了i的长度，要使得中位数med < mex，区间长度就不能超过 2*i，否则med 就会>= i了
以这一点为标准，就可以确定区间长度的上界了，然后下界就是最开始枚举出的l,r (必须包含0~i-1且无i)即最小区间。
由最小区间延伸的标准是：往 i 所在的方向延伸。
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int a[N], pos[N], n;
void solve () {
cin >> n;
for (int i = 0; i <= n; i++) pos[i] = 0;
for (int i = 1; i <= n; i++) cin >> a[i], pos[a[i]] = i;
long long ans = 0;
int l = pos[0], r = pos[0];
for (int i = 1; i <= n; i++) {
if (l <= pos[i] && pos[i] <= r) continue; //不满足mex
int len = r - l + 1;
if (len <= 2 * i) { //满足med
if (r < pos[i]) { //i在右
for (int j = r; j < pos[i]; j++) {
int dx = max (1, j - (2 * i - 1));
ans += max (0, l - dx + 1);
}
}
else {
for (int j = pos[i] + 1; j <= l; j++) {
int dx = min (n, j + (2 * i - 1));
ans += max (0, dx - r + 1);
}
}
}
l = min (l, pos[i]), r = max (r, pos[i]);
}
cout << ans << endl;
}
int main () {
int t;
cin >> t;
while (t --) solve ();
}
