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Niuke White Moon race 52
2022-06-30 08:01:00 【Chao Tang】
Niuke Xiaobai Yue race52
Problem A
Classification directe discussion jugement,7Chaque fois que le point arrive,Huit heures1Points~5En retard.,Les autres absences.
Problem B
Simulation,Appuyez.10La règle du nombre de bits pour essayer de s'adapter.Attention quand vous voyez 5000 Peut être déplacé pour devenir 10000 .
Problem C
Cumul des préfixes et des intervalles de maintenance,Enfin, voyez quelle partie a le moins de temps d'accumulation,À ce stade, toutes les instructions sauf celle - ci sont incorrectes, ce qui est le plus souvent le cas.
Problem D
Tableau arborescent & STTableau
Une table ronde,Il faut donc qu'il atteigne2X.Vous ne pouvez prendre qu'un arc à la fois par la valeur de l'intention du sujet（Ou un cercle entier）,Énumérez les aliments à partir de chaque gâteau à tour de rôle,Utilisez deux points pour trouver quel gâteau vous pouvez manger avec assez,Et ensuiteSTLe tableau indique les valeurs maximales de crème dans cette section.Complexité temporelle O ( n log n ) O(n\log n) O(nlogn)
C'est trop,Mais en écrivant le problème, je me demande pourquoi,Il semble que cette énumération ne garantit pas que le Groupe de crème qui a le plus haut et le plus bas niveau de crème n'inclut pas d'autres crèmes plus élevées？Comme si je n'étais pas très clair.
Problem E
Au début, j'ai pensé à l'équilibre,J'ai copié un modèle etWAC'est,Parce que la contribution des éléments des groupes ajoutés plus tard n'a pas été prise en compte.
La solution positive au problème est：Deux points plus
Découvrez la contribution de la valeur de la demande à tous les nombres,Moins la contribution de la valeur de la demande à ce groupe.
Problem F
Sac à dos de groupe,Non..
Code：
Problem A
// Good Good Study, Day Day AC.
#include <iostream>
#include <stdio.h>
#include <cstdio>
#include <stdlib.h>
#include <string>
#include <string.h>
#include <cstring>
#include <math.h>
#include <cmath>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <map>
#include <algorithm>
#include <unordered_map>
#include <unordered_set>
#define ffor(i,a,b) for(int i=(a) ;i<=(b) ;i++)
#define rrep(i,a,b) for(int i=(a) ;i>=(b) ;i--)
#define mst(v,s) memset(v,s,sizeof(v))
#define IOS ios::sync_with_stdio(false),cin.tie(0)
#define ll long long
#define INF 0x7f7f7f7f7f7f7f7f
#define inf 0x7f7f7f7f
#define PII pair<int,int>
#define int long long
using namespace std;
int n, T = 1;
void ready()
{
int a = 0, b = 0;
cin >> n;
while (n--) {
string st;
cin >> st;
if (st[0] == '7') continue;
if (st[0] == '9') b++;
if (st[0] == '8') {
int t = (st[2] - '0') * 10 + st[3] - '0';
if (t <= 5 && t != 0) a++;
else if (t > 5) b++;
}
}
cout << a << ' ' << b << '\n';
}
void work()
{
}
signed main()
{
IOS;
cin>>T;
while (T--) {
ready();
work();
}
return 0;
}
Problem B
// Good Good Study, Day Day AC.
#include <iostream>
#include <stdio.h>
#include <cstdio>
#include <stdlib.h>
#include <string>
#include <string.h>
#include <cstring>
#include <math.h>
#include <cmath>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <map>
#include <algorithm>
#include <unordered_map>
#include <unordered_set>
#define ffor(i,a,b) for(int i=(a) ;i<=(b) ;i++)
#define rrep(i,a,b) for(int i=(a) ;i>=(b) ;i--)
#define mst(v,s) memset(v,s,sizeof(v))
#define IOS ios::sync_with_stdio(false),cin.tie(0)
#define ll long long
#define INF 0x7f7f7f7f7f7f7f7f
#define inf 0x7f7f7f7f
#define PII pair<int,int>
#define int long long
using namespace std;
int n, T = 1;
void ready()
{
cin >> n;
}
void work()
{
int t = 1, ans = n;
if (ans < 10) {
if (ans >= 5) ans = 10;
}
while (n >= t) {
t *= 10;
int b = n % t;
if (b >= 10) while (b >= 10) b = b / 10;
n = n / t * t;
if (b >= 5) n += t;
//cout << "b=" << b << '\n';
ans = max(ans, n);
//cout << "n= " <<n<< '\n';
//cout << "t=" << t << '\n'<<'\n';
}
cout << ans << '\n';
}
signed main()
{
IOS;
cin>>T;
while (T--) {
ready();
work();
}
return 0;
}
Problem C
// Good Good Study, Day Day AC.
#include <iostream>
#include <stdio.h>
#include <cstdio>
#include <stdlib.h>
#include <string>
#include <string.h>
#include <cstring>
#include <math.h>
#include <cmath>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <map>
#include <algorithm>
#include <unordered_map>
#include <unordered_set>
#define ffor(i,a,b) for(int i=(a) ;i<=(b) ;i++)
#define rrep(i,a,b) for(int i=(a) ;i>=(b) ;i--)
#define mst(v,s) memset(v,s,sizeof(v))
#define IOS ios::sync_with_stdio(false),cin.tie(0)
#define ll long long
#define INF 0x7f7f7f7f7f7f7f7f
#define inf 0x7f7f7f7f
#define PII pair<int,int>
#define int long long
using namespace std;
const int N = 1e6 + 6;
int n, T = 1, m;
int sum[N];
void ready()
{
cin >> n >> m;
ffor(i,1,m) {
int op, x, y;
cin >> op;
if (op == 1) {
cin >> x >> y;
sum[x]++; sum[y + 1]--;
}
if (op == 2) {
cin >> x;
sum[x]++;
}
if (op == 3) {
cin >> x;
sum[1]++; sum[x+1]--;
}
}
int ans = inf, cnt = 0;
ffor(i, 1, n) {
sum[i] += sum[i - 1];
ans = min(sum[i], ans);
}
// ffor(i, 1, n) cout << sum[i] << ' '; cout << '\n';
cout << m - ans << ' ';
ffor(i, 1, n) {
if (sum[i] == ans) {
cnt++;
// cout << "i=" << i << '\n';
}
}
cout << cnt;
}
void work()
{
}
signed main()
{
IOS;
// cin>>T;
while (T--) {
ready();
work();
}
return 0;
}
Problem D
// Good Good Study, Day Day AC.
#include <iostream>
#include <stdio.h>
#include <cstdio>
#include <stdlib.h>
#include <string>
#include <string.h>
#include <cstring>
#include <math.h>
#include <cmath>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <map>
#include <algorithm>
#include <unordered_map>
#include <unordered_set>
#define ffor(i,a,b) for(int i=(a) ;i<=(b) ;i++)
#define rrep(i,a,b) for(int i=(a) ;i>=(b) ;i--)
#define mst(v,s) memset(v,s,sizeof(v))
#define IOS ios::sync_with_stdio(false),cin.tie(0)
#define ll long long
#define INF 0x7f7f7f7f7f7f7f7f
#define inf 0x7f7f7f7f
#define PII pair<int,int>
#define int long long
using namespace std;
const int N = 1e6 + 5;;
int n, T = 1, s;
int b[N], d[N][22];
int lowbite(int x)
{
return x & (-x);
}
void add_in(int i, int num, int c[])
{
while (i <= n * 2)
{
c[i] += num;
i += lowbite(i);
}
return;
}
int get_sum(int x, int c[])
{
int sum = 0;
while (x)
{
sum += c[x];
x -= lowbite(x);
}
return sum;
}
void ready()
{
cin >> n >> s;
ffor(i, 1, n) {
int bi;
cin >> bi;
add_in(i, bi, b);
add_in(i + n, bi, b);
}
ffor(i, 1, n) {
int di;
cin >> di;
d[i][0] = d[n + i][0] = di;
}
ffor(j, 1, 19) {
ffor(i, 1, 2 * n) {
d[i][j] = max(d[i][j - 1], d[i + (1 << (j - 1))][j - 1]);
}
}
}
int get_max(int x, int y)
{
int mid = log2(y - x + 1);
return max(d[x][mid], d[y - (1 << mid) + 1][mid]);
}
void work()
{
if (get_sum(n, b) < s) {
cout << -1;
return;
}
int ans = inf;
for (int i = 1; i <= n; i++) {
int l = i, r = n + i - 1, mid;
while (l < r) {
mid = l + r >> 1;
if (get_sum(mid, b) - get_sum(i - 1, b) < s) l = mid + 1;
else r = mid;
}
ans = min(ans, get_max(i, l));
}
cout << ans;
}
signed main()
{
IOS;
// cin>>T;
while (T--) {
ready();
work();
}
return 0;
}
Problem E
// Good Good Study, Day Day AC.
#include <iostream>
#include <stdio.h>
#include <cstdio>
#include <stdlib.h>
#include <string>
#include <string.h>
#include <cstring>
#include <math.h>
#include <cmath>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <map>
#include <algorithm>
#include <unordered_map>
#include <unordered_set>
#define ffor(i,a,b) for(int i=(a) ;i<=(b) ;i++)
#define rrep(i,a,b) for(int i=(a) ;i>=(b) ;i--)
#define mst(v,s) memset(v,s,sizeof(v))
#define IOS ios::sync_with_stdio(false),cin.tie(0)
#define ll long long
#define INF 0x7f7f7f7f7f7f7f7f
#define inf 0x7f7f7f7f
#define PII pair<int,int>
#define int long long
using namespace std;
const int N = 1e6 + 6;
const int mod = 998244353;
int n, T = 1, k;
vector<int> ve[N], alls;
void ready()
{
cin >> n >> k;
ffor(i, 1, n) {
int s;
cin >> s;
ffor(j, 1, s) {
int x;
cin >> x;
alls.push_back(x);
ve[i].push_back(x);
}
sort(ve[i].begin(), ve[i].end());
}
sort(alls.begin(), alls.end());
}
void work()
{
int ans = 0;
ffor(i, 1, n) {
for (auto item : ve[i]) {
int x = k - item;
ans += (alls.end() - lower_bound(alls.begin(), alls.end(), x)) - (ve[i].end() - lower_bound(ve[i].begin(), ve[i].end(), x));
}
}
ans /= 2;
cout << ans % mod;
}
signed main()
{
IOS;
// cin>>T;
while (T--) {
ready();
work();
}
return 0;
}
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