Codeforces Round 878 (Div. 3)
A：ABC
A. Cipher Shifer
题意：在自身后面添加一个字母，但是不能添加自身
思路：找到第二个与自身相符的就再找
#include <bits/stdc++.h>
using namespace std;
const int MAX = 110;
char a[MAX];
void solve() {
int n;
cin >> n;
for (int i = 0; i < n; i++) {
cin >> a[i];
}
int num = 0;
for (int i = 1; i < n; i++) {
if (a[num] == a[i]) {
cout << a[num];
num = ++i;
}
}
cout << "\n";
}
int main() {
int t;
cin >> t;
while (t--) {
solve();
}
return 0;
}
B. Binary Cafe
题意：求一个数n用<2^k的二进制数的表示方法
思路：每个数都有一个独立的二进制来表示，每个二进制表示都可记作一种方案（空集是一种单独的方案）int>1e9所以k>32直接输出n+1就行。其他的再判断就行
#include <bits/stdc++.h>
using namespace std;
#define int long long
int qmi(int a, int b) {
int res = 1;
while (b) {
if (b & 1) {
res = res * a;
}
a *= a;
b >>= 1;
}
return res;
}
void solve() {
int n, k;
cin >> n >> k;
if (k >= 32) {
cout << n + 1 << "\n";
} else {
int res = qmi(2, k) - 1;
if (res > n) cout << n + 1 << "\n";
else {
cout << res + 1 << "\n";
}
}
}
signed main() {
int t;
cin >> t;
while (t--) {
solve();
}
return 0;
}
C. Ski Resort
题意：一共n天，要连续去k天，这k天中a[i]<=q
思路：用一个记录有多少天连续满足这个条件，然后如果cnt>=k那么急可以在这几天去
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int MAX = 2e5 + 10;
bool t[MAX];
int dp[MAX];
void solve() {
int n, k, q;
cin >> n >> k >> q;
for (int i = 1; i <= n; i++) {
int x;
cin >> x;
if (x <= q) {
t[i] = true;
} else t[i] = false;
}
int cnt = 0;
for (int i = 1; i <= n; i++) {
if (t[i]) cnt++;
else cnt = 0;
dp[i] = dp[i - 1];
if (cnt >= k) {
dp[i] += cnt - k + 1;
}
}
cout << dp[n] << "\n";
}
signed main() {
int t;
cin >> t;
while (t--) {
solve();
}
return 0;
}
D. Wooden Toy Festival
题意：有x,y,z三个数，要让abs(a[i]-x)+abs(a[i]-y)+abs(a[i]-z)最小
思路：二分答案，是否有满足大于当前二分答案的两倍的个数大于3（大致分成三份），如果有就在大于当前二分的值的区间，否则就左边找
#include <bits/stdc++.h>
using namespace std;
#define int long long
const int MAX = 2e5 + 10;
int a[MAX];
int n;
bool check(int k) {
int num = 0;
int x = a[1];
for (int i = 1; i <= n; i++)
if (abs(a[i] - x) > k * 2) {
x = a[i];
num++;
}
if (num >= 3) return false;
return true;
}
void solve() {
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> a[i];
}
if (n <= 3) {
cout << "0" << endl;
return;
}
sort(a + 1, a + n + 1);
int l = 0, r = a[n];
int mid;
while (l <= r) {
mid = (l + r) >> 1;
if (check(mid)) r = mid - 1;
else l = mid + 1;
}
cout << l << "\n";
}
signed main() {
int t;
cin >> t;
while (t--) {
solve();
}
return 0;
}
E. Character Blocking
总结：大致思路有，但是没想到用map来存麻痹时间，题刷少了
题意：有三种操作:1.在 t 秒内屏蔽两个字符串中位于 pos位置的字符
2.交换两个未屏蔽的字符（注：不一定是不同字符串的）
3.确定两个字符在查询是是否相等
思路：找一个cnt确定有几个数是坏(没被屏蔽并且不相等)，用map来储存屏蔽时间
#include <bits/stdc++.h>
using namespace std;
void solve() {
string a, b;
cin >> a >> b;
a = " " + a;
b = " " + b;
int t, m;
cin >> t >> m;
int cnt = 0;
map<int, vector<int>> mp;
for (int i = 1; i < a.size(); i++) {
cnt += a[i] != b[i];
}
for (int i = 1; i <= m; i++) {//屏蔽时间问题
for (auto &x: mp[i]) {
if (a[x] != b[x]) cnt++;
}
int op;
cin >> op;
if (op == 1) {
int x;
cin >> x;
cnt -= (a[x] != b[x]);
mp[i + t].push_back(x);//i+t是解除屏蔽的时间
} else if (op == 2) {
int x1, y1, x2, y2;
cin >> x1 >> y1 >> x2 >> y2;
if (x1 == 1 && x2 == 1) {
if (a[y1] != b[y1]) cnt--;
if (a[y2] != b[y2]) cnt--;
swap(a[y1], a[y2]);
if (a[y1] != b[y1]) cnt++;
if (a[y2] != b[y2]) cnt++;
} else if (x1 == 1 && x2 == 2) {
if (a[y1] != b[y1]) cnt--;
if (a[y2] != b[y2]) cnt--;
swap(a[y1], b[y2]);
if (a[y1] != b[y1]) cnt++;
if (a[y2] != b[y2]) cnt++;
} else if (x1 == 2 && x2 == 1) {
if (b[y1] != a[y1]) cnt--;
if (b[y2] != a[y2]) cnt--;
swap(b[y1], a[y2]);
if (b[y1] != a[y1]) cnt++;
if (b[y2] != a[y2]) cnt++;
} else if (x1 == 2 && x2 == 2) {
if (b[y1] != a[y1]) cnt--;
if (b[y2] != a[y2]) cnt--;
swap(b[y1], b[y2]);
if (b[y1] != a[y1]) cnt++;
if (b[y2] != a[y2]) cnt++;
}
} else {
if (cnt) cout << "NO\n";
else cout << "YES\n";
}
}
}
int main() {
int t;
cin >> t;
while (t--) {
solve();
}
return 0;
}