Codeforces Round #828 (Div. 3) A~E泛做
挺有质量一套题
A.Number Replacement
对于数字串中出现次数\(\geq2\)的数字，判断是否对应同一个字母，出现次数为\(1\)的数字没有贡献。
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define cerr(x) std::cerr << (#x) << " is " << (x) << '\n'
#define IOS std::ios::sync_with_stdio(false);std::cin.tie(nullptr);
#define PII pair<int, int>
#define pdd pair<double,double>
#define LL long long
const double CLOCKS_PER_SECOND = ((clock_t) 1000);
const double CLOCKS_PER_MILLISECOND = ((clock_t) 1);
const int N = 2e5, M = 1e8, mod = 1e9 + 7, inf = 0x3f3f3f3f;
const double eps = 1e-6;
//#define x first
//#define y second
int T;
void solve() {
int n, a[50 + 5], ans1 = 0, ans2;
string s;
cin >> n;
map<int,char> temp;
map<int, int> m;
map<char, int> m2;
for (int i = 1; i <= n; i++) {
cin >> a[i];
m[a[i]]++;
}
cin >> s;
for (int i = 0; i < s.length(); i++)
m2[s[i]]++;
for (int i = 1; i <= n; i++) {
if (m[a[i]] >= 2) {
if (temp[a[i]] == 0) {
temp[a[i]] = s[i-1];
continue;
}
if (s[i-1] != temp[a[i]]) {
puts("NO");
return;
}
}
}
puts("YES");
return;
}
signed main() {
IOS;
cin >> T;
while (T--) {
solve();
}
return 0;
}
B.Even-Odd Increments
维护\(sum1\)和\(sum2\)，分别表示奇数的个数和偶数的个数。
判断每个\(x\)的奇偶性，然后讨论\(sum1\)和\(sum2\)的变化即可。
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define cerr(x) std::cerr << (#x) << " is " << (x) << '\n'
#define IOS std::ios::sync_with_stdio(false);std::cin.tie(nullptr);
#define PII pair<int, int>
#define pdd pair<double,double>
#define LL long long
const double CLOCKS_PER_SECOND = ((clock_t) 1000);
const double CLOCKS_PER_MILLISECOND = ((clock_t) 1);
const int N = 1e5, M = 1e8, mod = 1e9 + 7, inf = 0x3f3f3f3f;
const double eps = 1e-6;
//#define x first
//#define y second
int T;
void solve() {
int n, q;
int a[N];
cin >> n >> q;
LL sum = 0;
LL sum1 = 0, sum2 = 0;
for (int i = 1; i <= n; i++) {
cin >> a[i], sum += a[i];
if (a[i] & 1) sum1++; else sum2++;
}
while (q--) {
int opt;
LL x;
cin >> opt >> x;
if (opt == 0) {
if (x & 1LL) {
sum += sum2 * x;
sum1 = n;
sum2 = 0;
}//偶+奇=奇
else {
sum += sum2 * x;
}//偶+偶=偶
} else {
if (x & 1LL) {
sum += sum1 * x;
sum1 = 0;
sum2 = n;
}//奇+奇=偶
else {
sum += sum1 * x;
}//奇+偶=奇
}
cout << sum << endl;
}
}
signed main() {
IOS;
cin >> T;
while (T--) {
solve();
}
return 0;
}
C.Traffic Light
断环为链以后就是个\(O(2n)\)的暴力。
用\(string\)不知道为什么\(RE\)了，改成\(char\)才过。
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define cerr(x) std::cerr << (#x) << " is " << (x) << '\n'
#define IOS std::ios::sync_with_stdio(false);std::cin.tie(nullptr);
#define PII pair<int, int>
#define pdd pair<double,double>
#define LL long long
const double CLOCKS_PER_SECOND = ((clock_t) 1000);
const double CLOCKS_PER_MILLISECOND = ((clock_t) 1);
const int N = 2e5 + 10, M = 1e8, mod = 1e9 + 7, inf = 0x3f3f3f3f;
const double eps = 1e-6;
//#define x first
//#define y second
int T;
void solve() {
int n;
char ch;
char s[2 * N];
cin >> n >> ch;
cin >> s;
if (n == 1) {
cout << "0" << endl;
return;
}
int ans = 0;
int temp = 0;
for (int i = n; i <= 2 * n - 1; i++) {
s[i] = s[i - n];
}
for (int i = 2 * n - 1; i >= 0; i--) {
if (s[i] == 'g') temp = i;
else if (s[i] == ch) ans = max(ans, temp - i);
}
cout << ans << endl;
return;
}
signed main() {
IOS;
cin >> T;
while (T--) {
solve();
}
return 0;
}
D.Divisibility by 2^n
对\(a[i]\)质因数分解，统计\(2\)的个数\(t\)。如果\(t\geq n\)那么不需要执行操作。
否则统计每个\(i\)中质因数\(2\)的个数\(m_i\)，对\(i\)执行操作等价于\(t+=m_i\)
所以简单贪心一下就知道最少执行多少操作。
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define cerr(x) std::cerr << (#x) << " is " << (x) << '\n'
#define IOS std::ios::sync_with_stdio(false);std::cin.tie(nullptr);
#define PII pair<int, int>
#define pdd pair<double,double>
#define LL long long
const double CLOCKS_PER_SECOND = ((clock_t) 1000);
const double CLOCKS_PER_MILLISECOND = ((clock_t) 1);
const int N = 2e5 + 10, M = 1e8, mod = 1e9 + 7, inf = 0x3f3f3f3f;
const double eps = 1e-6;
//#define x first
//#define y second
int T;
void solve() {
int n, ans = 0, res = 0;
cin >> n;
int a[N];
vector<int> v;
for (int i = 1; i <= n; i++)
cin >> a[i];
for (int i = 1; i <= n; i++) {
int t = a[i], sum = 0;
if (t & 1) continue;
while (t % 2 == 0) t >>= 1, sum++;
res += sum;
}
if (res >= n) {
cout << 0 << endl;
return;
}
for (int i = 1; i <= n; i++) {
int t = i, sum = 0;
if (t & 1) continue;
while (t % 2 == 0) t >>= 1, sum++;
v.push_back(sum);
}
ans = res;
sort(v.begin(), v.end());
reverse(v.begin(), v.end());
for (int i = 0; i < v.size(); i++) {
ans += v[i];
if (ans >= n) {
cout << i + 1 << endl;
return;
}
}
cout << -1 << endl;
return;
}
signed main() {
IOS;
cin >> T;
while (T--) {
solve();
}
return 0;
}
E1. Divisible Numbers (easy version)
\(ab|xy\)等价于\(gcd(ab,xy)=ab\)。枚举\(x\)，那么\(y=\frac{ab}{gcd(ab,x)}\)。
令\(y=d/y*y\)，此时\(y\)一定小于等于\(d\)。只需要判断\(y>b\)即可。
\(O(c)\)
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define cerr(x) std::cerr << (#x) << " is " << (x) << '\n'
#define IOS std::ios::sync_with_stdio(false);std::cin.tie(nullptr);
#define PII pair<int, int>
#define pdd pair<double,double>
#define LL long long
const double CLOCKS_PER_SECOND = ((clock_t) 1000);
const double CLOCKS_PER_MILLISECOND = ((clock_t) 1);
const int N = 2e5 + 10, M = 1e8, mod = 1e9 + 7, inf = 0x3f3f3f3f;
const double eps = 1e-6;
//#define x first
//#define y second
int T;
LL _gcd(LL a, LL b) { return b == 0 ? a : _gcd(b, a % b); };
void solve() {
LL a, b, c, d;
cin >> a >> b >> c >> d;
for (LL x = a + 1; x <= c; x++) {
LL z = _gcd(a * b, x);
LL y = a * b / z;
y = d / y * y;
if (y > b) {
cout << x << " " << y << endl;
return;
}
}
cout << -1 << " " << -1 << endl;
return;
}
signed main() {
IOS;
cin >> T;
while (T--) {
solve();
}
return 0;
}
E2.Divisible Numbers (hard version)
参考@此处
显然枚举\(x\)是不可行的。我们发现\(xy\)是\(ab\)的倍数，我们尝试对\(a\)和\(b\)质因数分解，\(dfs\)穷举两两的因数积。将因数之积作为\(x\)，剩下的与\(E1\)一致。
一个\(longlong\)上限的数，约数数量级也仅是\(1e6\)的级别，所以对本题而言爆搜是完全可行的。
#include <bits/stdc++.h>
using namespace std;
#define endl '\n'
#define cerr(x) std::cerr << (#x) << " is " << (x) << '\n'
#define IOS std::ios::sync_with_stdio(false);std::cin.tie(nullptr);
#define PII pair<int, int>
#define pdd pair<double,double>
#define PLL pair<LL,LL>
#define LL long long
const double CLOCKS_PER_SECOND = ((clock_t) 1000);
const double CLOCKS_PER_MILLISECOND = ((clock_t) 1);
const int N = 2e5 + 10, M = 1e8, mod = 1e9 + 7, inf = 0x3f3f3f3f;
const double eps = 1e-6;
//#define x first
//#define y second
int T;
int prime[N], tot;
bool st[N];
LL _gcd(LL a, LL b) { return b == 0 ? a : _gcd(b, a % b); };
void init() {
for (int i = 2; i < N; i++) {
if (!st[i]) prime[tot++] = i;
for (int j = 0; prime[j] * i < N; j++) {
st[prime[j] * i] = 1;
if (!(i % prime[j])) break;
}
}
}
void solve1(LL x, vector<PLL > &v) {//对x质因数分解
for (int i = 0; i < tot; i++) {
if (x % prime[i]) continue;
int sum = 0;
while (!(x % prime[i])) x /= prime[i], sum++;
v.push_back({prime[i], sum});
}
if (x != 1) v.push_back({x, 1});
}
void dfs(int p, LL now, vector<PLL > &v, vector<LL> &v2) {
if (p == v.size()) {
v2.push_back(now);
return;
}
LL res = 1;
for (int i = 0; i <= v[p].second; i++) {
dfs(p + 1, now * res, v, v2);
if (i < v[p].second) res *= v[p].first;
}
}
void solve() {
LL a, b, c, d;
vector<PLL > v;
vector<LL> v2;
cin >> a >> b >> c >> d;
solve1(a, v);
solve1(b, v);
dfs(0, 1, v, v2);
sort(v2.begin(), v2.end());
for (int i = 0; i < v2.size(); i++) {
LL x = v2[i], y = a * b / x;
x = c / x * x, y = d / y * y;
if (x > a && y > b) {
cout << x << " " << y << endl;
return;
}
}
cout << -1 << " " << -1 << endl;
return;
}
signed main() {
IOS;
cin >> T;
init();
while (T--) {
solve();
}
return 0;
}
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