Codeforces Round 913 (Div. 3)（A~G）
1、编程模拟
2、栈模拟
3、找规律？（从终止状态思考）
4、二分
5、找规律，数学题
6、贪心（思维题）
7、基环树
A - Rook
题意：
直接模拟
// Problem: A. Rook
// Contest: Codeforces - Codeforces Round 913 (Div. 3)
// URL: https://codeforces.com/contest/1907/problem/0
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second
#define endl '\n'
const LL maxn = 4e05+7;
const LL N = 5e05+10;
const LL mod = 1e09+7;
const int inf = 0x3f3f3f3f;
const LL llinf = 5e18;
typedef pairpl;
priority_queue, greater >mi;//小根堆
priority_queue ma;//大根堆
LL gcd(LL a, LL b){
return b > 0 ? gcd(b , a % b) : a;
}
LL lcm(LL a , LL b){
return a / gcd(a , b) * b;
}
int n , m;
vectora(N , 0);
void init(int n){
for(int i = 0 ; i <= n ; i ++){
a[i] = 0;
}
}
void solve()
{
string s;
cin >> s;
char c = s[0];
int k = s[1] - '0';
for(int i = 0 ; i < 8 ; i++){
char t = i + 'a';
if(t == c){
continue;
}
else{
cout << t <>t;
while(t--)
{
solve();
}
return 0;
}
B - YetnotherrokenKeoard
题意：给定键入顺序，但是其中‘B’和‘b’不再代表字母，‘B’代表删除前面第一个大写字母，‘b’代表删除前面第一个小写字母，求最终键入的字符串结果。
思路：‘第一个’自然想到了栈的后进后出原则，‘B’‘b’操作即是出栈操作。先不考虑最终结果，而是将其字母的键入时间放入栈当中。最后将栈中键入时间全部取出，再根据键入时间输出结果。
// Problem: B. YetnotherrokenKeoard
// Contest: Codeforces - Codeforces Round 913 (Div. 3)
// URL: https://codeforces.com/contest/1907/problem/B
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second
#define endl '\n'
const LL maxn = 4e05+7;
const LL N = 5e05+10;
const LL mod = 1e09+7;
const int inf = 0x3f3f3f3f;
const LL llinf = 5e18;
typedef pairpl;
priority_queue, greater >mi;//小根堆
priority_queue ma;//大根堆
LL gcd(LL a, LL b){
return b > 0 ? gcd(b , a % b) : a;
}
LL lcm(LL a , LL b){
return a / gcd(a , b) * b;
}
int n , m;
vectora(N , 0);
void init(int n){
for(int i = 0 ; i <= n ; i ++){
a[i] = 0;
}
}
void solve()
{
string s;
cin >> s;
stackbig , small;
int len = s.size();
for(int i = 0 ; i < len ; i ++){
if(s[i] >= 'a' && s[i] <= 'z'){
if(s[i] == 'b'){
if(!small.empty()){
small.pop();
}
}
else{
small.push(i);
}
}
else{
if(s[i] == 'B'){
if(!big.empty()){
big.pop();
}
}
else{
big.push(i);
}
}
}
vectorans;
while(!big.empty()){
ans.pb(big.top());
big.pop();
}
while(!small.empty()){
ans.pb(small.top());
small.pop();
}
sort(ans.begin() , ans.end());
len = ans.size();
for(int i = 0 ; i < len ; i ++){
cout <>t;
while(t--)
{
solve();
}
return 0;
}
C - Removal of Unattractive Pairs
题意：给定字符串，能够删除相邻两个不相同字符，求最终字符最短能为多少。
思路：考虑最终状态：一定不存在字符或者只存在一种字符。因此考虑将其余字符全部删除，能将字符串变为多短。
// Problem: C. Removal of Unattractive Pairs
// Contest: Codeforces - Codeforces Round 913 (Div. 3)
// URL: https://codeforces.com/contest/1907/problem/C
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second
#define endl '\n'
const LL maxn = 4e05+7;
const LL N = 5e05+10;
const LL mod = 1e09+7;
const int inf = 0x3f3f3f3f;
const LL llinf = 5e18;
typedef pairpl;
priority_queue, greater >mi;//小根堆
priority_queue ma;//大根堆
LL gcd(LL a, LL b){
return b > 0 ? gcd(b , a % b) : a;
}
LL lcm(LL a , LL b){
return a / gcd(a , b) * b;
}
int n , m;
vectora(N , 0);
void init(int n){
for(int i = 0 ; i <= n ; i ++){
a[i] = 0;
}
}
void solve()
{
cin >> n;
vectorcnt(26 , 0);
string s;
cin >>s;
for(int i = 0; i < n ; i ++){
cnt[s[i] - 'a']++;
}
int maxx = 0;
for(int i = 0 ; i < 26 ; i ++){
maxx = max(maxx , cnt[i]);
}
cout << max( n & 1 ? 1 : 0 , n - 2 * (n - maxx)) << endl;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cout.precision(10);
int t=1;
cin>>t;
while(t--)
{
solve();
}
return 0;
}
D - Jumping Through Segments
题意：
思路：直接二分求答案，每一轮的范围是[上一轮左端点 - k ， 上一轮右端点 + k ] ,然后和当前线段合并，若不相交则false。
// Problem: D. Jumping Through Segments
// Contest: Codeforces - Codeforces Round 913 (Div. 3)
// URL: https://codeforces.com/contest/1907/problem/D
// Memory Limit: 256 MB
// Time Limit: 5000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second
#define endl '\n'
const LL maxn = 4e05+7;
const LL N = 5e05+10;
const LL mod = 1e09+7;
const int inf = 0x3f3f3f3f;
const LL llinf = 5e18;
typedef pairpl;
priority_queue, greater >mi;//小根堆
priority_queue ma;//大根堆
LL gcd(LL a, LL b){
return b > 0 ? gcd(b , a % b) : a;
}
LL lcm(LL a , LL b){
return a / gcd(a , b) * b;
}
int n , m;
vectora(N , 0);
void init(int n){
for(int i = 0 ; i <= n ; i ++){
a[i] = 0;
}
}
vector>range;
bool check(int len){
pair ran = {0 , 0};
int t = range.size();
for(auto to : range){
// cout << ran.x << " " << ran.y << endl;
int l = to.x;
int r = to.y;
ran.x -= len;
ran.y += len;
if(ran.x > r || ran.y < l)
return false;
ran.x = max(ran.x , l);
ran.y = min(ran.y , r);
}
return true;
}
void solve()
{
cin >> n;
for(int i = 0 ; i < n ; i ++){
pairtmp;
cin >> tmp.x >> tmp.y;
range.pb(tmp);
}
int l = 0 , r = 1e9;
while(l < r){
int mid = (l + r) / 2;
if(check(mid)){
r = mid;
}
else{
l = mid + 1;
}
}
cout << l << endl;
range.clear();
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cout.precision(10);
int t=1;
cin>>t;
while(t--)
{
solve();
}
return 0;
}
E - Good Triples
题意：
思路:可以发现，若要满足条件，那么三个数是不能进位的，因此本题中数据的每一位是不会影响的。满足乘法原则，将每一位所能带来的贡献相乘即可。
// Problem: E. Good Triples
// Contest: Codeforces - Codeforces Round 913 (Div. 3)
// URL: https://codeforces.com/contest/1907/problem/E
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second
#define endl '\n'
const LL maxn = 4e05+7;
const LL N = 5e05+10;
const LL mod = 1e09+7;
const int inf = 0x3f3f3f3f;
const LL llinf = 5e18;
typedef pairpl;
priority_queue, greater >mi;//小根堆
priority_queue ma;//大根堆
LL gcd(LL a, LL b){
return b > 0 ? gcd(b , a % b) : a;
}
LL lcm(LL a , LL b){
return a / gcd(a , b) * b;
}
int n , m;
vectora(N , 0);
void init(int n){
for(int i = 0 ; i <= n ; i ++){
a[i] = 0;
}
}
int dig(int s){
int cnt = 0;
while(s){
cnt += s % 10;
s/=10;
}
return cnt;
}
void solve()
{
// 0 == 1
// 1 == 3
// 2 == 6
// 3 == 10
// 4 == 15
// 5 == 21
// 6 == 28
// 7 == 36
// 8 == 45
// 9 == 55
// 3141 = 3 * 15 * 3 * 10
LL mask[10] = {1 , 3 , 6 , 10 , 15 , 21 , 28 , 36 , 45 , 55};
string s;
cin >> s;
int len = s.size();
LL ans = 1;
for(int i = 0 ; i < len ; i ++){
ans *= mask[s[i] - '0'];
}
cout << ans << endl;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cout.precision(10);
int t=1;
cin>>t;
while(t--)
{
solve();
}
return 0;
}
F - Shift and Reverse
思路：将数组放到环上考虑， 假设环起初按顺时针排 ， 且有一个指针指向了环的第一位。现考虑操作变成了什么：
移位操作：将指针往前移动一格。
反转：将指针往前一格，且访问顺序调换方向。
接下来考虑题目意思：即从环上某一点出发，绕一圈是非递减排序。其中指针位置可以变，指针访问方向也可以变，接下来就是一道模拟题了，模拟出最小操作数即可。
// Problem: F. Shift and Reverse
// Contest: Codeforces - Codeforces Round 913 (Div. 3)
// URL: https://codeforces.com/contest/1907/problem/F
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second
#define endl '\n'
const LL maxn = 4e05+7;
const LL N = 5e05+10;
const LL mod = 1e09+7;
const int inf = 0x3f3f3f3f;
const LL llinf = 5e18;
typedef pairpl;
priority_queue, greater >mi;//小根堆
priority_queue ma;//大根堆
LL gcd(LL a, LL b){
return b > 0 ? gcd(b , a % b) : a;
}
LL lcm(LL a , LL b){
return a / gcd(a , b) * b;
}
int n , m;
vectora(N , 0);
void init(int n){
for(int i = 0 ; i <= n ; i ++){
a[i] = 0;
}
}
void solve()
{
cin >> n;
for(int i = 0 ; i < n ; i ++)
cin >> a[i];
vectorv(2 * n , 0);
for(int i = 0 ; i < n ; i ++){
v[i] = a[i];
v[i + n] = a[i];
}
int ans = -1;
//递减
for(int i = 0 ; i < n ; ){
int st = i;
int cnt = 1;
while(i < 2 * n - 1 && v[i] >= v[i + 1]){
cnt++;
i++;
}
if(cnt >= n){
ans = min(1 + st, 1 + (n - st));
}
i ++;
}
//递增
for(int i = 0 ; i < n ; ){
int st = i;
int cnt = 1;
while(i < 2 * n - 1 && v[i] <= v[i + 1]){
cnt++;
i++;
}
if(cnt >= n){
if(ans == -1){
ans = min(n - st , st + 2);
}
else{
ans = min(ans ,min(n - st , st + 2) );
}
if(st == 0)
ans = 0;
}
i ++;
}
cout << ans << endl;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cout.precision(10);
int t=1;
cin>>t;
while(t--)
{
solve();
}
return 0;
}
G - Lights
题意：
思路：将相关联的灯建边，每次操作都对边两端处理。可以发现这是一颗基环树，也就是说：在非环上的点可以按照拓扑序来唯一的去处理（要么1变0，同时另一端也变化，要么不变）。现在就只剩下环上的一部分了，可以发现若环上剩余亮着的灯是奇数的话那么就不能全部熄灭了。
接下来考虑输出最小操作：若环上某一条边的操作已经确定下来了（变或者不变），那么其余所有边操作都确定了，因此可以考虑类似于环形dp的思想。将所有边操作分为两类表示对一类中的每一条边操作后整个环都能变成0。最后比较两类边集大小考虑输出哪一类即可。
// Problem: G. Lights
// Contest: Codeforces - Codeforces Round 913 (Div. 3)
// URL: https://codeforces.com/contest/1907/problem/G
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include
using namespace std;
#define LL long long
#define pb push_back
#define x first
#define y second
#define endl '\n'
const LL maxn = 4e05+7;
const LL N = 5e05+10;
const LL mod = 1e09+7;
const int inf = 0x3f3f3f3f;
const LL llinf = 5e18;
typedef pairpl;
priority_queue, greater >mi;//小根堆
priority_queue ma;//大根堆
LL gcd(LL a, LL b){
return b > 0 ? gcd(b , a % b) : a;
}
LL lcm(LL a , LL b){
return a / gcd(a , b) * b;
}
int n , m;
vectora(N , 0);
void init(int n){
for(int i = 0 ; i <= n ; i ++){
a[i] = 0;
}
}
void solve()
{
cin >> n;
vectorin(n + 5 , 0);
vectorvis(n + 5 , 0);
vectore(n + 5 , 0) ;
string s;
cin >> s;
s = " " + s;
for(int i = 1 ; i <= n ; i ++){
int u = i , v;
cin >> v;
e[u] = v;
in[v]++;
}
queueq;
for(int i = 1 ; i <= n ;i ++){
if(in[i] == 0){
q.push(i);
}
}
vectorans;
while(!q.empty()){
int x = q.front();
q.pop();
if(s[x] == '1'){
s[x] = '0';
ans.pb(x);
if(s[e[x]] == '1'){
s[e[x]] = '0';
}
else{
s[e[x]] = '1';
}
}
in[e[x]]--;
if(in[e[x]] == 0){
q.push(e[x]);
}
}
for(int i = 1 ; i <= n ; i ++){
if(in[i]){
int j = i;
int t = 0;//分为两组
int len = 0;
int res = 0;//t = 1 的组需要操作的数
while(in[j]){
if(s[j] == '1'){
t ^= 1;
}
res += t;
in[j] = 0;
len ++;
j = e[j];
}
if(t == 1){
cout << "-1\n";
return;
}
for(int k = 0 ; k < len ; k ++){
if(s[j] == '1'){
t ^= 1;
}
if(t == (res < len - res)){
ans.pb(j);
}
j = e[j];
}
}
}
cout << ans.size() << endl;
for(auto it : ans){
cout << it <<" ";
}
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cout.precision(10);
int t=1;
cin>>t;
while(t--)
{
solve();
}
return 0;
}