Codeforces Round #599 (Div. 2)
A. Maximum Square
-
思路：水题
-
AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
const int N = 1010;
int k, n;
int a[N];
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> k;
while (k -- ){
cin >> n;
int cnt = 0;
for (int i = 1; i <= n; i ++ )
cin >> a[i];
sort(a + 1, a + n + 1);
for (int i = n; i >= 1; i -- ){
if (cnt >= a[i])
break;
cnt ++ ;
}
cout << cnt << "
";
}
return 0;
}
B. Character Swap
-
思路：当a[i] != a[i]时 遍历是否存在a[j] == a[i] 若存在则交换a[j]和b[i] 否则判断是否存在a[i] == b[j] 若存在则交换a[j]和b[j] a[j]和b[i]
-
AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
int k, n;
bool flag;
string a, b;
vector<pair<int, int> > ans;
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> k;
while (k -- ){
flag = true;
ans.clear();
vector<int> cnt(26);
cin >> n >> a >> b;
a.insert(0, " "), b.insert(0, " ");
for (int i = 1; i <= n; i ++ ){
cnt[a[i] - 'a'] ++ ;
cnt[b[i] - 'a'] ++ ;
}
for (auto x: cnt){
if (x & 1){
flag = false;
break;
}
}
if (!flag){
cout << "No
";
continue;
}
cout << "Yes
";
for (int i = 1; i <= n; i ++ ){
if (a[i] != b[i]){
flag = false;
for (int j = i + 1; j <= n; j ++ ){
if (a[i] == a[j]){
flag = true;
ans.emplace_back(j, i);
swap(a[j], b[i]);
break;
}
}
if (!flag){
for (int j = i + 1; j <= n; j ++ ){
if (a[i] == b[j]){
flag = true;
ans.emplace_back(j, j);
ans.emplace_back(j, i);
swap(a[j], b[j]);
swap(a[j], b[i]);
break;
}
}
}
}
}
cout << ans.size() << "
";
for (auto x: ans)
cout << x.first << " " << x.second << "
";
}
return 0;
}
C. Tile Painting
-
思路：唯一分解定理后 对每一个p求个gcd即可
-
AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
const int N = 1e6 + 10;
ll n, ans;
vector<ll> p;
void divide(ll x){
ll i;
for (i = 2; i * i < x; i ++ ){
if (x % i == 0){
p.push_back(i);
p.push_back(x / i);
}
}
if (i * i == x)
p.push_back(i);
}
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n;
divide(n);
p.push_back(n);
ans = p[0];
for (int i = 0; i < p.size(); i ++ )
ans = gcd(ans, p[i]);
cout << ans << "
";
return 0;
}
D. 0-1 MST
-
思路：参考其他博客的思路 bfs暴力 求出连通块的个数 答案就是连通块的个数 - 1
-
AC代码
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <sstream>
#include <stack>
#include <stdio.h>
#include <string.h>
#include <string>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;
ll mult_mod(ll x, ll y, ll mod){
return (x * y - (ll)(x / (long double)mod * y + 1e-3) * mod + mod) % mod;
}
ll pow_mod(ll a, ll b, ll p){
ll res = 1;
while (b){
if (b & 1)
res = mult_mod(res, a, p);
a = mult_mod(a, a, p);
b >>= 1;
}
return res % p;
}
ll gcd(ll a, ll b){
return b ? gcd(b, a % b) : a;
}
const int N = 1e5 + 10;
int n, m, ans;
int vis[N];
set<int> st, G[N];
void bfs(int x){
queue<int> q;
q.push(x);
st.erase(x);
while (!q.empty()){
int y = q.front();
q.pop();
if (vis[y])
continue;
vis[y] = 1;
for (auto it = st.begin(); it != st.end(); ){
int v = *it;
++ it;
if (G[y].find(v) == G[y].end()){
q.push(v);
st.erase(v);
}
}
}
}
int main(){
#ifndef ONLINE_JUDGE
freopen("my_in.txt", "r", stdin);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n >> m;
for (int i = 1; i <= n; i ++ )
st.insert(i);
for (int i = 1; i <= m; i ++ ){
int x, y;
cin >> x >> y;
G[x].insert(y);
G[y].insert(x);
}
for (int i = 1; i <= n; i ++ ){
if (!vis[i]){
bfs(i);
ans ++ ;
}
}
cout << ans - 1 << "
";
return 0;
}