Educational Codeforces Round 80 题解

Educational Codeforces Round 80

比赛链接：

Educational Codeforces Round 80

A-Deadline

由 ​ ， ​得 ​.

所以我们只需要枚举​到 ​即可。

 #include 
 using namespace std;
 
 
define ll long long
 
 
const int maxn = 3e5+7;
 ll p[maxn];
 int main(){
 ios_base::sync_with_stdio(0);
 int t;
 cin >> t;
 while(t--){
 int n, d;
 cin >> n >> d;
 int ma = sqrt(d) + 10, i;
 for (i = 0; i < ma; i++){
 if(i + (d+i)/(i+1) <= n)
 break;
 }
 cout << (i < ma ? "YES" : "NO") << endl;
 }
 return 0;
 }

B - Yet Another Meme Problem

由题：

​

​

​

​

答案是 ​

 #include 
 using namespace std;
 
 
define ll long long
 
 
const int maxn = 3e5+7;
 ll p[maxn];
 int main(){
 ios_base::sync_with_stdio(0);
 int t;
 cin >> t;
 while(t--){
 ll a, b, cnt=0;
 cin >> a >> b;
 b++;
 while(b){
 cnt++;
 b /= 10;
 }
 ll ans = a * (cnt - 1);
 cout << ans << endl;
 }
 return 0;
 }

C - Two Arrays

由题目知道 ：

这是一个非递减序列。

这题有个有趣的数学知识，用到了搁板法，意思就是从个数中取个有多少种取法。

还有++必须用逆元来做，否则会输出.

用其他语言应该可以直接求C。

 #include 
 using namespace std;
 
 
define ll long long
 
 
const int mod = 1e9 + 7;
 const int maxn = 10005;
 ll fac[maxn];
 ​
 ​
 ll qpow(ll a, ll b){
 ll res = 1;
 while(b){
 if(b&1){
 res = (res * a) % mod;
 }
 a = (a * a) % mod;
 b >>= 1;
 }
 return res;
 }
 ll C(ll m, ll n){
 if(m>n || m <0)
 return 0;
 ll f1 = fac[n], f2 = fac[m] * fac[n - m] % mod;
 return f1 * qpow(f2, mod - 2) % mod;
 }
 int main(){
 ios_base::sync_with_stdio(0);
 fac[0] = 1;
 for (int i = 1; i < maxn; i++)
 fac[i] = fac[i - 1] * i % mod;
 ll n, m;
 cin >> n >> m;
 ll ans = C(n - 1, 2 * m + n - 1) % mod;
 cout << ans << endl;
 return 0;
 }const int N = 1e3 + 7, M = 13;
 int n, m;
 modint f[M][N], ans;
 ​
 int main() {
 rd(n), rd(m);
 f[0][1] = 1;
 for (int i = 0; i < m; i++)
 for (int j = 1; j <= n; j++)
 for (int k = j; k <= n; k++)
 f[i+1][k] += f[i][j];
 for (int i = 1; i <= n; i++)
 for (int j = i; j <= n; j++)
 ans += f[m][i] * f[m][n-j+1];
 print(ans);
 return 0;
 }

D - Minimax Problem

二分。

题目的讲解卿学姐实在是良心。

 #include 
 using namespace std;
 
 
define ll long long
 
 
const int maxn = 3e5+7;
 int n, m, ans_l, ans_r;
 int num[1000];
 int a[maxn][11];
 ​
 bool judge(int mid){
 memset(num, 0, sizeof(num));
 for (int i = 0; i < n; i++){
 int x = 0;
 for (int j = 0; j < m; j++){
 if(a[i][j] >= mid)
 x += (1 << j);
 }
 num[x] = i + 1;
 }
 for (int i = 0; i < (1 << m); i++){
 for (int j = 0; j < (1 << m); j++){
 if(num[i] && num[j] && ((i|j) == ((1< ans_l = num[i];
 ans_r = num[j];
 return true;
 }
 }
 }
 return false;
 }
 ​
 ​
 int main(){
 ios_base::sync_with_stdio(0);
 ​
 cin >> n >> m;
 int l = -1, r = 1e9;
 for (int i = 0; i < n; i++){
 for (int j = 0; j < m; j++){
 cin >> a[i][j];
 l = min(l, a[i][j]);
 r = max(r, a[i][j]);
 }
 }
 ​
 int ans = l;
 while(l<=r){
 int mid = (l + r) >> 1;
 if(judge(mid)){
 ans = mid;
 l = mid + 1;
 }
 else
 r = mid - 1;
 }
 judge(ans);
 cout << ans_l << " " << ans_r << endl;
 return 0;
 }

E - Messenger Simulator

树状数组。

 #include 
 using namespace std;
 
 
define ll long long
 
 
const int maxn = 1e6+7;
 ​
 int bit[maxn], a[maxn], max_a[maxn], min_a[maxn];
 int pos[maxn];
 int n, m;
 ​
 int lowbit(int x){
 return x & (-x);
 }
 ​
 void update(int x, int op){
 while(x < maxn){
 bit[x] += op;
 x += lowbit(x);
 }
 }
 int get(int x){
 int ans=0;
 while(x){
 ans += bit[x];
 x -= lowbit(x);
 }
 return ans;
 }
 int main(){
 ios_base::sync_with_stdio(0);
 cin >> n >> m;
 memset(bit, 0, sizeof(bit));
 for (int i = 1; i<=n; i++){
 max_a[i] = min_a[i] = i;
 pos[i] = i + m;
 update(i+m, 1);
 }
 for (int i = 0; i int x;
 cin >> x;
 min_a[x] = 1;
 max_a[x] = max(max_a[x], get(pos[x]));
 ​
 update(pos[x], -1);
 pos[x] = m - i;
 update(pos[x], 1);
 }
 for (int i = 1; i<=n; i++){
 max_a[i] = max(max_a[i], get(pos[i]));
 }
 for (int i = 1; i <= n; i++){
 cout << min_a[i] << " " << max_a[i] << endl;
 }
 return 0;
 }