Codeforces 920 反图联通块 线段树质因数暴力
A
#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que; get min
const double eps = 1.0e-10;
const double EPS = 1.0e-4;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
//const int maxn = 3e5 + 10;
const int turn[][] = {{, }, { -, }, {, }, {, -}};
//priority_queue<int, vector<int>, less<int>> que;
//next_permutation
ll Mod = ;
int num[];
int pre[];
int n;
int number;
bool check()
{
for (int i = ; i <= n; i++)
{
pre[i] = pre[i - ] + pre[i];
if (pre[i] < )
{
return false;
}
}
return true;
}
int main()
{
int T;
cin >> T;
while (T--)
{
cin >> n >> number;
for (int i = ; i <= number; i++)
{
cin >> num[i];
}
for (int i = ; i <= ; i++)
{
for (int j = ; j <= n; j++)
{
pre[j] = ;
}
for (int j = ; j <= number; j++)
{
if (i == )
{
pre[num[j]] += ;
pre[num[j] + ] += -;
}
else
{
pre[max(, num[j] - i + )] += ;
pre[num[j] + i] += -;
}
}
if (check())
{
cout << i << endl;
break;
}
}
}
return ;
}
B
#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que; get min
const double eps = 1.0e-8;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
//const int maxn = 3e5 + 10;
const int maxm = ;
const int turn[][] = {{, }, { -, }, {, }, {, -}};
//priority_queue<int, vector<int>, less<int>> que;
//next_permutation
ll mod = 3e7;
int le[];
int re[];
int wait[];
int main()
{
int t;
cin >> t;
while (t--)
{
int n;
scanf("%d", &n);
for (int i = ; i < n; i++)
{
scanf("%d %d", &le[i], &re[i]);
}
int cur = le[];
for (int i = ; i < n; i++)
{
cur = max(cur, le[i]);
if (re[i] >= cur)
{
wait[i] = cur;
cur++;
}
else
{
wait[i] = ;
}
}
for (int i = ; i < n; i++)
{
cout << wait[i];
if (i != n - )
{
cout << " ";
}
}
cout << endl;
}
return ;
}
C
#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que; get min
const double eps = 1.0e-10;
const double EPS = 1.0e-4;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
//const int maxn = 3e5 + 10;
const int turn[][] = {{, }, { -, }, {, }, {, -}};
//priority_queue<int, vector<int>, less<int>> que;
//next_permutation
ll Mod = ;
int num[];
int pre[];
char f[];
int main()
{
int n;
cin >> n;
for (int i = ; i <= n; i++)
{
scanf("%d", num + i);
}
scanf("%s", f + );
for (int i = ; i < n; i++)
{
if (f[i] == '')
{
pre[i] = pre[i - ] + ;
}
else
{
pre[i] = ;
}
}
for (int i = ; i < n; i++)
{
if (pre[i] > pre[i + ])
{
sort(num + i - pre[i] + , num + i + );
}
}
for (int i = ; i < n; i++)
{
if (num[i] != i)
{
cout << "NO" << endl;
return ;
}
}
cout << "YES" << endl;
return ;
}
E
把所有点放在一个set里，每次取set中一个顶点，删去，遍历set，删去与此顶点邻接的顶点
因为遍历的过程中有两种结局1.删去某个结点 遍历成功 2.两点之间不存在边 遍历失败
所以遍历的总复杂度为O(n+m) 再加上set的复杂度就是 O((n+m)log)
#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que; get min
const double eps = 1.0e-10;
const double EPS = 1.0e-4;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
//const int maxn = 3e5 + 10;
const int turn[][] = {{, }, { -, }, {, }, {, -}};
//priority_queue<int, vector<int>, less<int>> que;
//next_permutation
ll Mod = ;
set<int> need;
map<int, bool> mp[];
vector<int> ans;
int main()
{
int n, m;
int from, to;
cin >> n;
for (int i = ; i <= n; i++)
{
need.insert(i);
}
cin >> m;
for (int i = ; i <= m; i++)
{
scanf("%d %d", &from, &to);
mp[from][to] = mp[to][from] = ;
}
while (!need.empty())
{
int todo = *need.begin();
need.erase(todo);
queue<int> que;
ans.push_back();
que.push(todo);
while (!que.empty())
{
queue<int> shan;
int cnt = que.front();
que.pop();
ans.back()++;
for (auto i : need)
{
if (!mp[cnt][i])
{
que.push(i);
shan.push(i);
}
}
while (!shan.empty())
{
need.erase(shan.front());
shan.pop();
}
}
}
cout << ans.size() << endl;
sort(ans.begin(), ans.end());
for (auto i : ans)
{
cout << i << " ";
}
cout << endl;
return ;
}
F
如果知道到N的因数(N%i==0)数量级是N^(1/3)的这道题就很好做了 因为当N=1或者N=2时因数数目等于N 而1e6=2^20 每个数最多被修改7次
所以线段树维护一个最大值 一个sum值 当最大值不大于2时不用修改 大于二时递归下去暴力修改
#include <bits/stdc++.h>
#define PI acos(-1.0)
#define mem(a,b) memset((a),b,sizeof(a))
#define TS printf("!!!\n")
#define pb push_back
#define inf 1e9
//std::ios::sync_with_stdio(false);
using namespace std;
//priority_queue<int,vector<int>,greater<int>> que; get min
const double eps = 1.0e-10;
const double EPS = 1.0e-4;
typedef pair<int, int> pairint;
typedef long long ll;
typedef unsigned long long ull;
//const int maxn = 3e5 + 10;
const int turn[][] = {{, }, { -, }, {, }, {, -}};
//priority_queue<int, vector<int>, less<int>> que;
//next_permutation
ll Mod = ;
ll n, q;
int l, r;
ll dp[];
struct node
{
ll maxn, sum;
} tree[];
void pushup(int x)
{
tree[x].sum = tree[x << ].sum + tree[x << | ].sum;
tree[x].maxn = max(tree[x << ].maxn, tree[x << | ].maxn);
}
void build(int x, ll value, int root = , int l = , int r = n)
{
if (l > x || r < x)
{
return ;
}
if (l == x && r == x)
{
tree[root].sum = tree[root].maxn = value;
return;
}
int mid = (l + r) >> ;
if (x <= mid)
{
build(x, value, root << , l, mid);
}
else
{
build(x, value, root << | , mid + , r);
}
pushup(root);
}
void update(int xl, int xr, int root = , int l = , int r = n)
{
if (l > r || l > xr || r < xl)
{
return;
}
if (xl <= l && xr >= r && tree[root].maxn <= )
{
return;
}
if (l == r)
{
tree[root].sum = tree[root].maxn = dp[tree[root].sum];
return ;
}
int mid = (l + r) >> ;
if (xl <= mid)
{
update(xl, xr, root << , l, mid);
}
if (xr > mid)
{
update(xl, xr, root << | , mid + , r);
}
pushup(root);
}
ll getsum(int xl, int xr, int root = , int l = , int r = n)
{
if (l > r || l > xr || r < xl)
{
return ;
}
if (xl <= l && xr >= r)
{
return tree[root].sum;
}
int mid = (l + r) >> ;
return getsum(xl, xr, root << , l, mid) + getsum(xl, xr, root << | , mid + , r);
}
int main()
{
cin >> n >> q;
ll cnt;
for (int i = ; i <= ; i++)
{
for (int j = i; j <= ; j += i)
{
dp[j]++;
}
}
for (int i = ; i <= n; i++)
{
scanf("%lld", &cnt);
build(i, cnt);
}
for (int i = ; i <= q; i++)
{
int now;
ll value;
int aim;
cin >> now;
if (now == )
{
scanf("%d %d", &l, &r);
update(l, r);
}
else
{
scanf("%d %d", &l, &r);
cout << getsum(l, r) << endl;
}
}
return ;
}
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