\[
Time Limit: 1 s\quad Memory Limit: 256 MB
\]
取模计算即可。
view
/***************************************************************
> File Name : a.cpp
> Author : Jiaaaaaaaqi
> Created Time : 2020/1/4 19:57:41
***************************************************************/
#include <map>
#include <set>
#include <list>
#include <ctime>
#include <cmath>
#include <stack>
#include <queue>
#include <cfloat>
#include <string>
#include <vector>
#include <cstdio>
#include <bitset>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
#define lowbit(x) x & (-x)
#define mes(a, b) memset(a, b, sizeof a)
#define fi first
#define se second
#define pb push_back
#define pii pair<int, int>
typedef unsigned long long int ull;
typedef long long int ll;
const int maxn = 1e5 + 10;
const int maxm = 1e5 + 10;
const ll mod = 1e9 + 7;
const ll INF = 1e18 + 100;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-8;
using namespace std;
int n, m;
int cas, tol, T;
char s[50][50], t[50][50];
int main() {
// freopen("in", "r", stdin);
scanf("%d%d", &n, &m);
for(int i=1; i<=n; i++) scanf("%s", s[i]+1);
for(int i=1; i<=m; i++) scanf("%s", t[i]+1);
scanf("%d", &T);
while(T--) {
ll x;
scanf("%lld", &x);
ll a = x%n, b = x%m;
if(a==0) a = n;
if(b==0) b = m;
printf("%s%s\n", s[a]+1, t[b]+1);
}
return 0;
}\[
Time Limit: 2 s\quad Memory Limit: 256 MB
\]
对于每一个数列，先判断其本身是否可以拿出两个递增的数。如果可以的话，其可以和任意其他数列组成一个合法答案。如果不可以的话，可以尝试把其放在前面，选出他的最小值，然后找出有多少个数列可以放在后面，并且选出数列的最大值大于此数列的最小值，这一段可以通过二分得到。
剩下的只要扣除重复计算的部分即可。
view
/***************************************************************
> File Name : b.cpp
> Author : Jiaaaaaaaqi
> Created Time : 2020/1/4 20:19:01
***************************************************************/
#include <map>
#include <set>
#include <list>
#include <ctime>
#include <cmath>
#include <stack>
#include <queue>
#include <cfloat>
#include <string>
#include <vector>
#include <cstdio>
#include <bitset>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
#define lowbit(x) x & (-x)
#define mes(a, b) memset(a, b, sizeof a)
#define fi first
#define se second
#define pb push_back
#define pii pair<int, int>
typedef unsigned long long int ull;
typedef long long int ll;
const int maxn = 1e5 + 10;
const int maxm = 1e5 + 10;
const ll mod = 1e9 + 7;
const ll INF = 1e18 + 100;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-8;
using namespace std;
int n, m;
int cas, tol, T;
bool vis[maxn];
int a[maxn], b[maxn], c[maxn];
vector<int> g;
int main() {
// freopen("in", "r", stdin);
ll ans = 0;
scanf("%d", &n);
g.clear();
int cnt = 0;
for(int i=1; i<=n; i++) {
scanf("%d", &m);
a[i] = inf, b[i] = 0;
for(int j=1; j<=m; j++) {
scanf("%d", &c[j]);
a[i] = min(a[i], c[j]);
b[i] = max(b[i], c[j]);
}
vis[i] = 0;
int Min = c[1];
for(int j=2; j<=m; j++) {
if(Min < c[j]) {
vis[i] = 1;
break;
}
Min = min(Min, c[j]);
}
if(!vis[i]) g.pb(b[i]);
if(vis[i]) ans += n+n-1, cnt++;
}
ans -= 1ll*(cnt)*(cnt-1);
sort(g.begin(), g.end());
g.pb(inf);
for(int i=1; i<=n; i++) {
if(vis[i]) continue;
int p = upper_bound(g.begin(), g.end(), a[i])-g.begin();
ans += g.size()-p-1;
}
printf("%lld\n", ans);
return 0;
}\[
Time Limit: 1 s\quad Memory Limit: 256 MB
\]
枚举区间长度，然后计算可以拿多少个连续的区间放在多少段连续的位置，然后分成选出的数和没选出的数全排列放即可。
view
/***************************************************************
> File Name : c.cpp
> Author : Jiaaaaaaaqi
> Created Time : 2020/1/4 21:03:22
***************************************************************/
#include <map>
#include <set>
#include <list>
#include <ctime>
#include <cmath>
#include <stack>
#include <queue>
#include <cfloat>
#include <string>
#include <vector>
#include <cstdio>
#include <bitset>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
#define lowbit(x) x & (-x)
#define mes(a, b) memset(a, b, sizeof a)
#define fi first
#define se second
#define pb push_back
#define pii pair<int, int>
typedef unsigned long long int ull;
typedef long long int ll;
const int maxn = 1e6 + 10;
const int maxm = 1e5 + 10;
const ll INF = 1e18 + 100;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-8;
using namespace std;
ll n, m, mod;
int cas, tol, T;
ll A[maxn];
int main() {
// freopen("in", "r", stdin);
scanf("%lld%lld", &n, &mod);
A[0] = 1%mod;
for(int i=1; i<=n; i++) A[i] = A[i-1]*i%mod;
ll ans = 0;
for(int i=1; i<=n; i++) {
ans += A[i]*A[n-i]%mod*(n-i+1)%mod*(n-i+1)%mod;
ans %= mod;
}
printf("%lld\n", ans);
return 0;
}\[
Time Limit: 2 s\quad Memory Limit: 256 MB
\]
对于每个时间点 \(t\)，我们可以维护有哪些会议 \(t\) 时刻正在 \(A\) 场地进行，假设有 \(cnt\) 个会议，那么这 \(cnt\) 个会议都是两两冲突的，那么这 \(cnt\) 个会议在 \(B\) 场地也必须是两两冲突的，也就是说必须有一个时刻被这些会议的 \([sb, eb]\) 区间覆盖。
那么可以在 \(sa\) 时刻加入 \([sb, eb]\) 区间，在 \(ea+1\) 时刻删除 \([sb, eb]\) 区间，那么可以用线段树来维护每一时刻在 \(B\) 场地被多少个会议覆盖，然后就可以通过线段树的最大值来判断是否合法。
此时判断的是 \(A\) 场地冲突时，\(B\) 场地是不是都冲突。只需要再反过来判断 \(B\) 场地冲突时，\(A\) 场地是不是都冲突即可。
view
/***************************************************************
> File Name : d.cpp
> Author : Jiaaaaaaaqi
> Created Time : 2020/1/6 21:03:02
***************************************************************/
#include <map>
#include <set>
#include <list>
#include <ctime>
#include <cmath>
#include <stack>
#include <queue>
#include <cfloat>
#include <string>
#include <vector>
#include <cstdio>
#include <bitset>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
#define lowbit(x) x & (-x)
#define mes(a, b) memset(a, b, sizeof a)
#define fi first
#define se second
#define pb push_back
#define pii pair<int, int>
typedef unsigned long long int ull;
typedef long long int ll;
const int maxn = 4e5 + 10;
const int maxm = 1e5 + 10;
const ll mod = 1e9 + 7;
const ll INF = 1e18 + 100;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-8;
using namespace std;
int n, m;
int cas, tol, T;
vector<int> vv;
vector<pii> g[maxn];
int sa[maxn], ea[maxn], sb[maxn], eb[maxn];
int node[maxn<<2], lazy[maxn<<2];
int getid(int x) {
return lower_bound(vv.begin(), vv.end(), x) - vv.begin()+1;
}
void build(int l, int r, int rt) {
node[rt] = lazy[rt] = 0;
if(l==r) return ;
int mid = l+r>>1;
build(l, mid, rt<<1);
build(mid+1, r, rt<<1|1);
}
void pushdown(int rt) {
if(lazy[rt]) {
lazy[rt<<1] += lazy[rt];
lazy[rt<<1|1] += lazy[rt];
node[rt<<1] += lazy[rt];
node[rt<<1|1] += lazy[rt];
lazy[rt] = 0;
}
}
void update(int l, int r, int pl, int pr, int val, int rt) {
if(pl<=l && r<=pr) {
node[rt] += val, lazy[rt] += val;
return ;
}
int mid = l+r>>1;
pushdown(rt);
if(pl<=mid) update(l, mid, pl, pr, val, rt<<1);
if(pr>mid) update(mid+1, r, pl, pr, val, rt<<1|1);
node[rt] = max(node[rt<<1], node[rt<<1|1]);
}
bool ok() {
int cnt = 0;
build(1, tol, 1);
for(int i=1; i<=tol; i++) {
for(auto t : g[i]) update(1, tol, sb[t.se], eb[t.se], t.fi, 1), cnt += t.fi;
if(node[1] != cnt) return false;
}
return true;
}
int main() {
// freopen("in", "r", stdin);
scanf("%d", &n);
for(int i=1; i<=n; i++) {
scanf("%d%d%d%d", &sa[i], &ea[i], &sb[i], &eb[i]);
vv.pb(sa[i]), vv.pb(ea[i]), vv.pb(sb[i]), vv.pb(eb[i]);
}
sort(vv.begin(), vv.end()), vv.erase(unique(vv.begin(), vv.end()), vv.end());
for(int i=1; i<=n; i++) {
sa[i] = getid(sa[i]), ea[i] = getid(ea[i]);
sb[i] = getid(sb[i]), eb[i] = getid(eb[i]);
g[sa[i]].pb({1, i});
g[ea[i]+1].pb({-1, i});
}
tol = vv.size();
if(!ok()) return 0*puts("NO");
for(int i=1; i<=tol+1; i++) g[i].clear();
for(int i=1; i<=n; i++) {
swap(sa[i], sb[i]), swap(ea[i], eb[i]);
g[sa[i]].pb({1, i});
g[ea[i]+1].pb({-1, i});
}
if(!ok()) return 0*puts("NO");
return 0*puts("YES");
}原文：https://www.cnblogs.com/Jiaaaaaaaqi/p/12158877.html