A题直接枚举即可,枚举日期,暴力匹配
#include<iostream> #include<algorithm> #include<vector> using namespace std;
bool check(string t){ if(t.substr(0, 4)!="2023") return false; string mon = t.
substr(4, 2); string day = t.substr(6, 2); int m = (mon[0]-'0')*10+(mon[1]-'0');
int d = (day[0]-'0')*10+(day[1]-'0'); if(m==1||m==3||m==5||m==7||m==8||m==10||m
==12){ if(d>=1&&d<=31) return true; }else if(m==2){ if(d>=1&&d<=28) return true;
}else if(m==4||m==6||m==9||m==11){ if(d>=1&&d<=30) return true; } return false;
}; string p="",s=""; int ans = 0, n; string mon[]={"01","02","03","04","05","06"
,"07","08","09","10","11","12"}; string day[]={"01","02","03","04","05","06",
"07","08","09","10", "11","12","13","14","15","16","17","18","19","20", "21",
"22","23","24","25","26","27","28","29","30","31"}; int md[]={31,28,31,30,31,30,
31,31,30,31,30,31}; bool check(string t,string s){ int j=0; for(int i=0;i<t.size
();++i){ if(j==s.size()) return true; if(t[i]==s[j]) j++; } return j==s.size();
} void solve(){ while(cin >> s){ p += s; } // cout << p << endl; n=p.size();
string t= "2023"; for(int i=0;i<12;++i){ for(int j=0;j<md[i];++j){ string N = t
+ mon[i] + day[j]; if(check(p, N)) { ans ++; } } } cout<<ans<<endl; } int main()
{ solve(); return 0; }
答案:

可以看出香浓信息熵有单调性(在0不超过1这个前提下)
因此直接二分即可,顺便输出一下结果对应的函数值
#include<bits/stdc++.h> using namespace std; #define int long long double p(int
x,int y){ //x 0 y 1 double P0 = 1.0*x/(x+y); double P1 = 1.0*y/(x+y); return -
1.0*x*x/(x+y)*log2(P0)-1.0*y*y/(x+y)*log2(P1); } void solve(){ int n = 23333333;
int l=1, r=(n-1)/2; double PP = 11625907.5798; while(l<r){ int mid = (l+r)>>1;
double t = p(mid, n-mid); if(t<PP) l=mid+1; else r=mid; } cout << l << endl; l=
11027421; printf("%.5lf", p(l, n-l)); } signed main(){ solve(); return 0; }

貌似可以直接O(1)算,但是我选择直接二分

二分的正确性在保证有解的前提下成立
#include<iostream> #include<vector> using namespace std; //V:normal -> 1:X void
solve(){ int n; cin >> n; long long x,y,X=1e9,Y=0; vector<pair<int,int>> vec;
for(int i=0;i<n;++i){ cin >> x >> y; vec.push_back({x,y}); } auto check=[&](long
long x)->bool{ for(auto u:vec){ if(u.first/x < u.second) return false; } return
true; }; long long l=1,r=1e9; while(l<r){ long long mid=(l+r+1)>>1; if(check(mid
)) l=mid; // x/mid >= y else r=mid-1; } Y=l; auto find=[&](long long x)->bool{
for(auto u:vec){ if(u.first/x > u.second) return false; } return true; }; l=1,r=
1e9; while(l<r){ long long mid=(l+r)>>1; if(find(mid)) r=mid; else l=mid+1; } X=
l; cout<<X<<" " << Y<<endl; } int main(){ solve(); return 0; }
顺便跑了几组对拍:

N ≤ \le ≤ 10 直接搜,不要想什么贪心
#include<iostream> #include<vector> #include<array> using namespace std; bool
vis[20]; int n; bool dfs(int num,int now, vector<array<int,3>>& vec){ if(num==n)
{ return true; } int flag = 0; for(int i=0;i<n;++i){ if(!vis[i]){ if(now>vec[i][
0]+vec[i][1]) return false; vis[i] = true; flag |= dfs(num+1, now+vec[i][2], vec
); vis[i]=false; } } return flag; } void solve(){ cin >> n; for(int i=1;i<=n;++i
) vis[i] = 0; vector<array<int,3>> vec; for(int i=0;i<n;++i){ int x,y,z; cin >>
x>> y >> z; vec.push_back({x,y,z}); } bool res = dfs(0, 0, vec); cout<<(res?
"YES":"NO")<<"\n"; } int main(){ ios::sync_with_stdio(false); cin.tie(0); int _=
1; cin >>_; while(_--) solve(); return 0; } /* 2 3 0 100 10 10 10 10 0 2 20 3 0
10 20 10 10 20 20 10 20 */

删最少数使得他是接龙序列,即求原序列的最大接龙子序列,然后用n减去。
最大接龙子序列求法:DP
但是我们发现转移类型一共只有9种1~9结尾的数字
因此我们可以开一个数组f[i]表示前i个数字,以第i个结尾的最大接龙子序列。
再开一个数组bac[1~9]表示 以1~9结尾的最大接龙子序列
每次用第i个数组开头的字母对应的数组更新f[i],然后再用f[i]反过来更新bac即可
细节看代码
#include<iostream> #include<vector> #include<array> #include<string> using
namespace std; int n; void solve(){ cin >> n; vector<string> vec; for(int i=0;i<
n;++i){ string t; cin >> t; vec.push_back(t); } vector<int> bac(10); //1,2,3....
vector<int> f(n+1, 0); int ans = 0 ; for(int i=0;i<n;++i){ f[i]=max(1, bac[vec[
i][0]-'0']+1); ans=max(ans, f[i]); bac[vec[i].back()-'0'] = max(bac[vec[i].back(
)-'0'], f[i]); } cout << n-ans << endl; } int main(){ ios::sync_with_stdio(false
); cin.tie(0); int _=1; // cin >>_; while(_--) solve(); return 0; }

先用边缘的海水进行bfs求出"真的海水",然后再对所有岛屿进行BFS,此时BFS的时候非真的海水可以直接识别为陆地然后入队。
两遍BFS得到答案。
#include<iostream> #include<vector> #include<array> #include<string> #include
<queue> #define x first #define y second using namespace std; int n,m; char G[60
][60]; int vis[60][60], g[60][60], water[60][60]; int d[][2]={1,0,0,1,-1,0,0,-1}
; int mov[][2]={{1,0},{-1,0},{0,1},{0,-1},{1,-1}, {1,1},{-1,-1},{-1,1}}; void
solve(){ cin >> n >> m; for(int i=1;i<=n;++i){ for(int j=1;j<=m;++j){ vis[i][j]
= g[i][j] = water[i][j] = 0; } } for(int i=1;i<=n;++i){ cin >> (G[i]+1); for(int
j=1;j<=m;++j){ g[i][j] = G[i][j]-'0'; } } queue<pair<int,int>> q; for(int i=1;i
<=n;++i){ if(g[i][1] == 0){ water[i][1] = true; q.push({i, 1}); } if(g[i][m] ==
0){ water[i][m] = true; q.push({i, m}); } } for(int i=1;i<=m;++i){ if(g[1][i] ==
0){ if(!water[1][i]){ water[1][i] = true; q.push({1, i}); } } if(g[n][i] == 0){
if(!water[n][i]){ water[n][i] = true; q.push({n, i}); } } } while(q.size()){
auto u=q.front(); q.pop(); int x=u.x,y=u.y; for(int i=0;i<8;++i){ int xx=x+mov[i
][0], yy=y+mov[i][1]; if(xx>=1&&xx<=n&&yy>=1&&yy<=m&&!water[xx][yy]&&!g[xx][yy])
{ q.push({xx, yy}); water[xx][yy] = true; } } } while(q.size()) q.pop(); int idx
= 0; for(int i=1;i<=n;++i){ for(int j=1;j<=m;++j){ if(!water[i][j]&&g[i][j]&&!
vis[i][j]){ q.push({i, j}); vis[i][j]=++idx; while(q.size()){ auto u=q.front();
q.pop(); int x=u.x,y=u.y; for(int i=0;i<4;++i){ int xx=x+d[i][0], yy=y+d[i][1];
if(xx>=1&&xx<=n&&yy>=1&&yy<=m &&!vis[xx][yy]&&!water[xx][yy]){ vis[xx][yy]=idx;
q.push({xx,yy}); } } } } } } cout << idx << endl; } int main(){ ios::
sync_with_stdio(false); cin.tie(0); int _=1; cin >>_; while(_--) solve(); return
0; } /* 2 5 5 01111 11001 10101 10001 11111 5 6 111111 100001 010101 100001
111111 */

二分/双指针都行
先按照位置处理出来两个数组
然后枚举开头的位置,二分出结尾在另一个数组的合法位置,直接累加答案
#include<iostream> #include<vector> #include<array> #include<string> #include
<queue> #define x first #define y second using namespace std; int k; char st,ed;
string p; void solve(){ cin >> k; cin >> p >> st >> ed; vector<int> ps,pe; for(
int i=0;i<p.size();++i){ if(p[i]==st) ps.push_back(i); if(p[i]==ed) pe.push_back
(i); } long long ans = 0; for(int i=0;i<ps.size();++i){ int x = ps[i]; int X = x
+ k - 1; int l=0,r=pe.size()-1; while(l<r){ int mid=(l+r)>>1; if(pe[mid] >= X) r
=mid; else l=mid+1; } if(pe[l] >= X){ ans += pe.size() - l; } } cout<<ans<<endl;
} int main(){ ios::sync_with_stdio(false); cin.tie(0); int _=1; // cin >>_;
while(_--) solve(); return 0; }

双链表维护每个数左边和右边分别是什么数
set维护每个位置的值和要删的最小值
本题考察STL的用法
#include<iostream> #include<vector> #include<algorithm> #include<array> #
include<string> #include<queue> #include<set> #define x first #define y second
using namespace std; int n,k; void solve(){ cin >> n >> k; vector<int> a(n+1),
pre(n+1),nxt(n+1); set<pair<int,int>> ds; for(int i=1;i<=n;++i){ pre[i] = i-1;
nxt[i] = i+1; } for(int i=1;i<=n;++i){ cin >> a[i]; ds.insert({a[i], i}); } for(
int t=1;t<=k;++t){ auto pos=ds.begin(); int x=pos->first, y=pos->second; int l =
pre[y], r = nxt[y]; ds.erase(pos); pre[nxt[y]]=pre[y]; nxt[pre[y]]=nxt[y]; pre[
y]=-1, nxt[y]=-1; if(l>0){ ds.erase(ds.find({a[l], l})); a[l] += x; ds.insert({a
[l], l}); } if(r<=n){ ds.erase(ds.find({a[r], r})); a[r] += x; ds.insert({a[r],
r}); } } vector<pair<int,int>> temp; for(auto u:ds){ temp.push_back({u.y, u.x});
} sort(temp.begin(), temp.end()); for(auto u:temp){ cout << u.y << " "; }cout<<
"\n"; } int main(){ ios::sync_with_stdio(false); cin.tie(0); int _=1; // cin
>>_; while(_--) solve(); return 0; }

LCA, 没啥好说的
#include<iostream> #include<vector> #include<array> #include<string> #include
<queue> #define x first #define y second using namespace std; #define debug(x)
cout<<#x<<": "<<x<<endl #define N 100010 vector<pair<int,int>> edge[N]; int n,m,
k; int dep[N],fa[20][N]; int rout[N]; queue<int> q; void bfs(){ for(int i=1;i<=n
;++i) dep[i]=-1; dep[1]=0; q.push(1); while(q.size()){ auto u=q.front(); q.pop()
; for(auto v:edge[u]){ int ver = v.x, w=v.y; if(dep[ver]==-1){ dep[ver] = dep[u]
+ w; q.push(ver); fa[0][ver] = u; for(int i=1;i<=19;++i){ fa[i][ver] = fa[i-1][
fa[i-1][ver]]; } } } } } int lca(int x,int y){ if(dep[x] < dep[y]) swap(x, y);
for(int i=19;i>=0;--i){ if(dep[fa[i][x]] >= dep[y]) x = fa[i][x]; } if(x==y)
return x; for(int i=19;i>=0;--i){ if(fa[i][x]!=fa[i][y]) x=fa[i][x], y=fa[i][y];
} return fa[0][x]; } int dist(int x,int y){ return dep[x]+dep[y]-2*dep[lca(x,y)]
; } void solve(){ cin >> n >> k; for(int i=1;i<n;++i){ int x,y,z; cin >> x >> y
>> z; edge[x].push_back({y, z}); edge[y].push_back({x, z}); } bfs(); vector<int>
a(k+1); for(int i=1;i<=k;++i){ cin >> a[i]; } int now = a[1]; long long time = 0
; for(int i=1;i<=k;++i){ time += dist(now, a[i]); now = a[i]; } cout << time -
dist(a[1], a[2]) << " "; now = a[1]; for(int i=2;i<=k;++i){ int temp = time;
temp-= dist(now, a[i]); if(i!=k){ temp -= dist(a[i], a[i+1]); temp += dist(now,
a[i+1]); } cout << temp << " "; now = a[i]; }cout << "\n"; } int main(){ ios::
sync_with_stdio(false); cin.tie(0); int _=1; // cin >>_; while(_--) solve();
return 0; } /* 6 4 1 2 1 1 3 1 3 4 2 3 5 2 4 6 3 2 6 5 1 */

树上差分,也是LCA,最后再DFS一下
#include<iostream> #include<vector> #include<array> #include<string> #include
<queue> #define x first #define y second using namespace std; #define debug(x)
cout<<#x<<": "<<x<<endl #define N 100010 int n,m,k; vector<int> edge[N]; pair<
int,int> p[N]; int w[N],dep[N],fa[20][N]; pair<int,int> edges[N]; void bfs(){
for(int i=1;i<=n;++i) dep[i]=-1; queue<int> q; q.push(1); dep[1]=1; while(q.size
()){ auto u=q.front(); q.pop(); for(auto v:edge[u]){ if(dep[v]==-1){ dep[v]=dep[
u]+1; q.push(v); fa[0][v] = u; for(int i=1;i<=19;++i) fa[i][v]=fa[i-1][fa[i-1][v
]]; } } } } int lca(int x,int y){ if(dep[x]<dep[y]) swap(x, y); for(int i=19;i>=
0;--i){ if(dep[fa[i][x]] >= dep[y]) x=fa[i][x]; } if(x==y) return x; for(int i=
19;i>=0;--i){ if(fa[i][x]!=fa[i][y]) x=fa[i][x], y=fa[i][y]; } return fa[0][x];
} int f[N]; void dfs(int u,int pre){ f[u] = w[u]; for(auto v:edge[u]){ if(v!=pre
){ dfs(v, u); f[u] += f[v]; } } } void solve(){ cin >> n >> m; for(int i=1;i<n;
++i){ int x,y; cin >> x >> y; edge[x].push_back(y); edge[y].push_back(x); edges[
i] = {x, y}; } bfs(); for(int i=1;i<=m;++i){ cin >> p[i].x >> p[i].y; w[p[i].x]
++, w[p[i].y] ++; w[fa[0][lca(p[i].x, p[i].y)]] -= 1; w[lca(p[i].x, p[i].y)] -=
1; } dfs(1, -1); int ans = -1; for(int i=1;i<n;++i){ int x=edges[i].x, y=edges[i
].y; if(dep[x]<dep[y]) swap(x, y); //y-x if(f[x]==m){ ans=max(ans, i); } } cout
<<ans<<endl; } int main(){ ios::sync_with_stdio(false); cin.tie(0); int _=1; //
cin >>_; while(_--) solve(); return 0; }

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