A - delete .
在输出时过滤一遍即可
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| #include <bits/stdc++.h>
#define endl '\n'
using ll = long long;
using namespace std;
const ll N = 2e5 + 5;
int main() {
string s;
cin >> s;
string ans;
for (int i = 0; i < s.size(); i++) {
if (s[i] != '.') {
ans += s[i];
}
}
cout << ans << endl;
return 0;
}
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B - 3^A
打表后排序输出一遍即可
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| #include <bits/stdc++.h>
#define endl '\n'
using ll = long long;
using namespace std;
const ll N = 2e5 + 5;
vector<ll> q;
ll m;
bool cmp(ll a, ll b) {
return a > b;
}
void solve() {
int temp = 1;
while (temp <= m) {
q.push_back(temp);
temp *= 3;
}
}
int main() {
cin >> m;
solve();
sort(q.begin(), q.end(), cmp);
vector<ll> ans;
ll i = 0;
while (m > 0) {
while (q[i] <= m) {
ans.push_back(q.size() - i - 1);
m -= q[i];
}
i++;
}
cout << ans.size() << endl;
for (auto i : ans) {
cout << i << " ";
}
return 0;
}
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C - Count ABC Again
根据题意模拟一遍即可
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| #include <bits/stdc++.h>
#define endl '\n'
using ll = long long;
using namespace std;
const ll N = 2e5 + 5;
int main() {
int n, q;
cin >> n >> q;
string s;
cin >> s;
ll ans = 0;
for (int i = 0; i < n; i++) {
if (s[i] == 'A') {
if (s[i + 1] == 'B' && s[i + 2] == 'C') {
ans++;
}
}
}
while(q--) {
int pos;
string t;
cin >> pos >> t;
if (s[pos - 1] == t[0]) {
cout << ans << endl;
continue;
}
if (s[pos - 1] == 'A') {
if (s[pos] == 'B' && s[pos + 1] == 'C') {
if (t[0] != 'A') {
ans--;
}
}
}
if (s[pos - 1] == 'B') {
if (s[pos - 2] == 'A' && s[pos] == 'C') {
if (t[0] != 'B') {
ans--;
}
}
}
if (s[pos - 1] == 'C') {
if (s[pos - 2] == 'B' && s[pos - 3] == 'A') {
if (t[0] != 'C') {
ans--;
}
}
}
s[pos - 1] = t[0];
if (t[0] == 'A') {
if (s[pos] == 'B' && s[pos + 1] == 'C') {
ans++;
}
}
if (t[0] == 'B') {
if (s[pos - 2] == 'A' && s[pos] == 'C') {
ans++;
}
}
if (t[0] == 'C') {
if (s[pos - 2] == 'B' && s[pos - 3] == 'A') {
ans++;
}
}
cout << ans << endl;
}
return 0;
}
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D - Buildings
题目要求找到i之后有几个j可以使得a[i + 1 ….. j]中的所有值 <= a[j], 双循环写会超时，考虑枚举j，找到第一个i使得a[i] > a[j], 则i + 1 到 j的所以贡献都加1.
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| #include <bits/stdc++.h>
#define endl '\n'
using ll = long long;
using namespace std;
const ll N = 2e5 + 5;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n;
cin >> n;
vector<ll> a(n);
for (int i = 0; i < n; i++) {
cin >> a[i];
}
stack<ll> s;
vector<ll> ans(n);
for (int i = n - 1; i >= 0; i--) {
ans[i] = s.size();
while (s.size() && a[s.top()] < a[i]) {
s.pop();
}
s.push(i);
}
for (int i = 0; i < n; i++) {
cout << ans[i] << endl;
}
return 0;
}
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