2022ICPC南京

A:

#include<bits/stdc++.h>
using namespace std;
void solve() {
int n, m, k;
cin >> n >> m >> k;
vector<vector<int>> mp(n+10, vector<int>(m+10));
vector<vector<int>> vis(n+10, vector<int>(m+10));
string s;
cin >> s;
int L = 1, R = m;
int U = 1, D = n;
int curL = 1, curR = m;
int curU = 1, curD = n;
for (int i = 0; i < s.length(); i++) {
if (s[i] == 'L') {
curL++, curR++;
} else if (s[i] == 'R') {
curL--; curR--;
} else if (s[i] == 'U') {
curU++; curD++;
} else {
curU--; curD--;
}
L = max(curL, L);
R = min(curR, R);
U = max(curU, U);
D = min(curD, D);
}
if (U > D || L > R) {
if (k == 0) cout << n * m << 'n';
else cout << 0 << 'n';
return;
}
int xa = U, ya = L;
int xb = D, yb = R;
auto add = [&](int x1, int y1, int x2, int y2) {
if (vis[x1][y1]) return;
vis[x1][y1] = 1;
mp[x1][y1]++;
mp[x2+1][y1]--;
mp[x1][y2+1]--;
mp[x2+1][y2+1]++;
};
add(xa, ya, xb, yb);
for (int i = 0; i < s.length(); i++) {
if (s[i] == 'L') {
ya--; yb--;
} else if (s[i] == 'R') {
ya++; yb++;
} else if (s[i] == 'U') {
xa--; xb--;
} else {
xa++; xb++;
}
add(xa, ya, xb, yb);
}
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
mp[i][j] += mp[i-1][j] + mp[i][j-1] - mp[i-1][j-1];
//cout << mp[i][j] << ' ';
}
// cout << 'n';
}
int cnt = (D - U + 1) * (R - L + 1);
int ans = 0;
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= m; j++) {
if (cnt - mp[i][j] == k) ans++;
}
}
cout << ans << 'n';
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
int T;
cin >> T;
while (T--) solve();
return 0;
}

G

贪心策略，能选合并就合并，不能合并就选择加一只。
这里我们用了二分，直接二分最多需要多少次合并

#include<bits/stdc++.h>
using namespace std;
void solve() {
int n;
cin >> n;
int x = 1, y = 1;
vector<int> events(n);
for (int i = 0; i < n; i++) cin >> events[i];
auto check = [&](int pos, bool ans) -> bool {
int x = 1, y = 1;
for (int i = 0; i < n; i++) {
int op = events[i];
if (op == 0) {
if (pos <= i) op = -1;
else op = 1;
}
if (op == -1) {
if (y < 2) return false;
y--;
} else {
x++; y++;
}
}
if (ans) {
int g = __gcd(x, y);
x /= g; y /= g;
cout << x << ' ' << y << 'n';
}
return true;
};
int ans = n;
int l = 0, r = n-1;
while (l <= r) {
int mid = (l + r) >> 1;
if (check(mid, false)) {
ans = mid;
r = mid - 1;
} else {
l = mid + 1;
}
}
if (!check(ans, true)) cout << -1 << 'n';
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
int T;
cin >> T;
while (T--) solve();
return 0;
}

I

#include <iostream>
#include <vector>
using namespace std;
int main()
{
int T;
cin >> T;
while (T -- )
{
string s;
cin >> s;
vector<int> cnt(26, 0);
int res = 0;
for (auto &it : s) cnt[it - 'a'] ++, res = max(res, cnt[it - 'a']);
cout << s.size() - res << endl;
}
return 0;
}

最后

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