C题
题目大意：
现在给定一个字符串求解出当前序列大于等于A和小于B的子数组中的个数
思路：
可以使用二分查找，因为A需要维护大于等于，所以查找出当前位置大于等于A的位置，B需要维护小于，所以查找到当前小于B的位置，两个位置中间的位置都符合
代码：
cpp
#include<bits/stdc++.h>
#define int long long
//#define ull unsigned long long
using namespace std;
//#define x first
//#define y second
const int mod=19650827;
//const int N=1e5+10;
//int fac[N],invfac[N];
//int qmi(int a,int b,int p){int res=1;while(b){if(b&1)res=res*a%p;a=a*a%p;b>>=1;}return res;}
//int inv(int x){return qmi(x,p-2);}
//int C(int n,int m){if(n<0||m<0||n<m)return 0;return fac[n]*invfac[n-m]%p*invfac[m]%p;}
//void init(){fac[0]=1;for(int i=1;i<N;i++)fac[i]=fac[i-1]*i%p;invfac[N-1]=inv(fac[N-1]);for(int i=N-2;i>=0;i--)invfac[i]=invfac[i+1]*(i+1)%p;}
//struct node{int v,w;bool operator < (const node &x)const{return w==x.w ? v>x.v : w>x.w;}};
//void dijkstra(const vector<vector<node>>&g,vector<int>&dist,vector<bool>&vis,int x){fill(dist.begin(),dist.end(),LLONG_MAX);fill(vis.begin(),vis.end(),false);
//dist[x]=0;priority_queue<node>q;q.push({x,0});while(q.size()){auto u=q.top().v;q.pop();if(vis[u])continue;vis[u]=true;for(auto [v,w]:g[u]){if(dist[v]>dist[u]+w){
//dist[v]=dist[u]+w;q.push({v,dist[v]});}}}}
//struct BIT{int n;vector<int>t;BIT(int _n=0){init(_n);}void init(int _n){n=_n;t.assign(n+2,0);}
//void add(int x,int v){for(;x<=n;x+=(x & -x))t[x]+=v;}int query(int x){int res=0;for(;x;x-=(x & -x))res+=t[x];return res;}};
//int calc_add(const vector<vector<int>>&s,int x1,int y1,int x2,int y2){return s[x2][y2]-s[x2][y1-1]-s[x1-1][y2]+s[x1-1][y1-1];}
//void calc_diff(vector<vector<int>>&d,int x1,int y1,int x2,int y2,int c){d[x1][y1]+=c;d[x2+1][y1]-=c;d[x1][y2+1]-=c;d[x2+1][y2+1]+=c;}
//int primes[N],cnt=0;bool st[N];
//void init1(){for(int i=2;i<N;i++){if(!st[i])primes[++cnt]=i;for(int j=1;primes[j]*i<N;j++){st[primes[j]*i]=true;if(i%primes[j]==0)break;}}}
//vector<int>factor[N];void init2(){for(int i=1;i<N;i++){for(int j=i;j<N;j+=i){factor[j].push_back(i);}}}
//int find(vector<int>&pre,int x){return pre[x]=(pre[x]==x ? x : find(pre,pre[x]));}
//void merge(vector<int>&pre,int x,int y){int fx=find(pre,x),fy=find(pre,y);if(fx==fy)return;pre[fx]=fy;}
//bool check(vector<int>&pre,int x,int y){return find(pre,x)==find(pre,y);}
//string add(string x,string y){if((int)x.size()<(int)y.size())swap(x,y);int n=x.size();int m=y.size();reverse(x.begin(),x.end());reverse(y.begin(),y.end());
//int d=0;string ans;for(int i=0;i<n;i++){int x1=(x[i]-'0');int y1=(i<m ? y[i]-'0' : 0);int sum=x1+y1+d;ans.push_back((sum%10)+'0');d=sum/10;}
//if(d>0)ans.push_back(d+'0');reverse(ans.begin(),ans.end());return ans;}
//string mul(string x,string y){if(x=="0"||y=="0")return "0";reverse(x.begin(),x.end());reverse(y.begin(),y.end());int x1=x.size();int y1=y.size();string ans;
//for(int i=0;i<x1;i++){int d=0;int val1=(x[i]-'0');for(int j=0;j<y1;j++){int val2=(y[j]-'0');int len=ans.size();int val=val1*val2+d;
//if(len<=i+j)ans.push_back((val%10)+'0');else val+=(ans[i+j]-'0'),ans[i+j]=((val%10)+'0');d=val/10;}int len=ans.size();int cur_pos=i+y1;
//while(d){if(cur_pos>=len)ans.push_back((d%10)+'0');else d+=ans[cur_pos]-'0',ans[cur_pos]=((d%10)+'0');d/=10;cur_pos++;}}reverse(ans.begin(),ans.end());return ans;}
//struct Seg_Tree{int n;vector<int>t,lz;Seg_Tree(int _n=0){init(_n);}void init(int _n){n=_n;t.assign((n<<2)+2,0);lz.assign((n<<2)+2,0);}
//void pushup(int o){t[o]=t[o<<1]+t[o<<1|1];}void update(int s,int e,int o,int x){t[o]+=x*(e-s+1);lz[o]+=x;}
//void pushdown(int s,int e,int o){if(!lz[o])return ;int mid=(s+e)>>1;update(s,mid,o<<1,lz[o]);update(mid+1,e,o<<1|1,lz[o]);lz[o]=0;}
//void add(int l,int r,int x,int s=1,int e=0,int o=1){if(e==0)e=this->n;if(l>e||r<s)return ;if(l<=s&&e<=r){update(s,e,o,x);return ;}pushdown(s,e,o);int mid=(s+e)>>1;
//add(l,r,x,s,mid,o<<1);add(l,r,x,mid+1,e,o<<1|1);pushup(o);}
//int query(int l,int r,int s=1,int e=0,int o=1){if(e==0)e=this->n;if(l>e||r<s)return 0;if(l<=s&&r>=e)return t[o];pushdown(s,e,o);int mid=(s+e)>>1;
//int x=query(l,r,s,mid,o<<1);int y=query(l,r,mid+1,e,o<<1|1);return x+y;}};
void solve(){
int n,a,b;
cin>>n>>a>>b;
string s;
cin>>s;
s=' '+s;
vector<int>s1(n+1),s2(n+1);
for(int i=1;i<=n;i++){
if(s[i]=='a')s1[i]=s1[i-1]+1;
else s1[i]=s1[i-1];
if(s[i]=='b')s2[i]=s2[i-1]+1;
else s2[i]=s2[i-1];
}
int ans=0;
for(int l=1;l<=n;l++){
int r1=lower_bound(s1.begin()+l,s1.end(),a+s1[l-1])-s1.begin();
//if(r1>n)continue;
int r2=lower_bound(s2.begin()+l,s2.end(),b+s2[l-1])-s2.begin()-1;
//r2=min(r2,n);
if(r2>=r1)ans+=(r2-r1+1);
}
cout<<ans<<'\n';
}
signed main(){
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
//init();
//init1();
//init2();
int _=1;
//cin>>_;
while(_--)solve();
return 0;
}D题
题目大意：
给定一个序列，首先第一个人在0的位置，之后的人的位置依次给出，求解每一个人除了自己的位置以外距离当前点最近的和
思路：
可以发现这个题一定是一个在线处理的方法，将每一个点加入可以发现只与当前点的前驱和后继有关，求解当前点的前驱和后继可以使用平衡树
代码：
cpp
#include<bits/stdc++.h>
#define int long long
//#define ull unsigned long long
using namespace std;
//#define x first
//#define y second
//const int mod=19650827;
//const int N=1e5+10;
//int fac[N],invfac[N];
//int qmi(int a,int b,int p){int res=1;while(b){if(b&1)res=res*a%p;a=a*a%p;b>>=1;}return res;}
//int inv(int x){return qmi(x,p-2);}
//int C(int n,int m){if(n<0||m<0||n<m)return 0;return fac[n]*invfac[n-m]%p*invfac[m]%p;}
//void init(){fac[0]=1;for(int i=1;i<N;i++)fac[i]=fac[i-1]*i%p;invfac[N-1]=inv(fac[N-1]);for(int i=N-2;i>=0;i--)invfac[i]=invfac[i+1]*(i+1)%p;}
//struct node{int v,w;bool operator < (const node &x)const{return w==x.w ? v>x.v : w>x.w;}};
//void dijkstra(const vector<vector<node>>&g,vector<int>&dist,vector<bool>&vis,int x){fill(dist.begin(),dist.end(),LLONG_MAX);fill(vis.begin(),vis.end(),false);
//dist[x]=0;priority_queue<node>q;q.push({x,0});while(q.size()){auto u=q.top().v;q.pop();if(vis[u])continue;vis[u]=true;for(auto [v,w]:g[u]){if(dist[v]>dist[u]+w){
//dist[v]=dist[u]+w;q.push({v,dist[v]});}}}}
//struct BIT{int n;vector<int>t;BIT(int _n=0){init(_n);}void init(int _n){n=_n;t.assign(n+2,0);}
//void add(int x,int v){for(;x<=n;x+=(x & -x))t[x]+=v;}int query(int x){int res=0;for(;x;x-=(x & -x))res+=t[x];return res;}};
//int calc_add(const vector<vector<int>>&s,int x1,int y1,int x2,int y2){return s[x2][y2]-s[x2][y1-1]-s[x1-1][y2]+s[x1-1][y1-1];}
//void calc_diff(vector<vector<int>>&d,int x1,int y1,int x2,int y2,int c){d[x1][y1]+=c;d[x2+1][y1]-=c;d[x1][y2+1]-=c;d[x2+1][y2+1]+=c;}
//int primes[N],cnt=0;bool st[N];
//void init1(){for(int i=2;i<N;i++){if(!st[i])primes[++cnt]=i;for(int j=1;primes[j]*i<N;j++){st[primes[j]*i]=true;if(i%primes[j]==0)break;}}}
//vector<int>factor[N];void init2(){for(int i=1;i<N;i++){for(int j=i;j<N;j+=i){factor[j].push_back(i);}}}
//int find(vector<int>&pre,int x){return pre[x]=(pre[x]==x ? x : find(pre,pre[x]));}
//void merge(vector<int>&pre,int x,int y){int fx=find(pre,x),fy=find(pre,y);if(fx==fy)return;pre[fx]=fy;}
//bool check(vector<int>&pre,int x,int y){return find(pre,x)==find(pre,y);}
//string add(string x,string y){if((int)x.size()<(int)y.size())swap(x,y);int n=x.size();int m=y.size();reverse(x.begin(),x.end());reverse(y.begin(),y.end());
//int d=0;string ans;for(int i=0;i<n;i++){int x1=(x[i]-'0');int y1=(i<m ? y[i]-'0' : 0);int sum=x1+y1+d;ans.push_back((sum%10)+'0');d=sum/10;}
//if(d>0)ans.push_back(d+'0');reverse(ans.begin(),ans.end());return ans;}
//string mul(string x,string y){if(x=="0"||y=="0")return "0";reverse(x.begin(),x.end());reverse(y.begin(),y.end());int x1=x.size();int y1=y.size();string ans;
//for(int i=0;i<x1;i++){int d=0;int val1=(x[i]-'0');for(int j=0;j<y1;j++){int val2=(y[j]-'0');int len=ans.size();int val=val1*val2+d;
//if(len<=i+j)ans.push_back((val%10)+'0');else val+=(ans[i+j]-'0'),ans[i+j]=((val%10)+'0');d=val/10;}int len=ans.size();int cur_pos=i+y1;
//while(d){if(cur_pos>=len)ans.push_back((d%10)+'0');else d+=ans[cur_pos]-'0',ans[cur_pos]=((d%10)+'0');d/=10;cur_pos++;}}reverse(ans.begin(),ans.end());return ans;}
//struct Seg_Tree{int n;vector<int>t,lz;Seg_Tree(int _n=0){init(_n);}void init(int _n){n=_n;t.assign((n<<2)+2,0);lz.assign((n<<2)+2,0);}
//void pushup(int o){t[o]=t[o<<1]+t[o<<1|1];}void update(int s,int e,int o,int x){t[o]+=x*(e-s+1);lz[o]+=x;}
//void pushdown(int s,int e,int o){if(!lz[o])return ;int mid=(s+e)>>1;update(s,mid,o<<1,lz[o]);update(mid+1,e,o<<1|1,lz[o]);lz[o]=0;}
//void add(int l,int r,int x,int s=1,int e=0,int o=1){if(e==0)e=this->n;if(l>e||r<s)return ;if(l<=s&&e<=r){update(s,e,o,x);return ;}pushdown(s,e,o);int mid=(s+e)>>1;
//add(l,r,x,s,mid,o<<1);add(l,r,x,mid+1,e,o<<1|1);pushup(o);}
//int query(int l,int r,int s=1,int e=0,int o=1){if(e==0)e=this->n;if(l>e||r<s)return 0;if(l<=s&&r>=e)return t[o];pushdown(s,e,o);int mid=(s+e)>>1;
//int x=query(l,r,s,mid,o<<1);int y=query(l,r,mid+1,e,o<<1|1);return x+y;}};
void solve(){
int n;
cin>>n;
vector<int>x(n+2);
map<int,int>p;
for(int i=1;i<=n;i++)cin>>x[i];
set<int>s;
s.insert(0);
//需要维护当前点距离哪一个点最近
vector<int>y(n+2);
x[0]=0,y[0]=0,p[0]=0;
int ans=0;
for(int i=1;i<=n;i++){
p[x[i]]=i;
auto pre=s.lower_bound(x[i]);
pre--;
auto suf=s.upper_bound(x[i]);
int pre1=p[*pre];
int suf1=(suf!=s.end())? p[*suf] : 0;
//求解前缀的值要不要改变
if(i==1||abs(*pre-x[i])<abs(*pre-x[y[pre1]])){
ans-=abs(*pre-x[y[pre1]])-abs(*pre-x[i]);
y[pre1]=i;
}
//求解后缀的值要不要改变
if(suf!=s.end()&&abs(*suf-x[i])<abs(*suf-x[y[suf1]])){
ans-=abs(*suf-x[y[suf1]])-abs(*suf-x[i]);
y[suf1]=i;
}
//处理当前加入的点应该加入的距离
if(suf==s.end()||(abs(*suf-x[i])>abs(*pre-x[i]))){
ans+=abs(*pre-x[i]);
//当前的值距离后缀的值最近
y[i]=pre1;
}
else{
ans+=abs(*suf-x[i]);
//当前的值距离前缀的值最近
y[i]=suf1;
}
s.insert(x[i]);
cout<<ans<<'\n';
}
}
signed main(){
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
//init();
//init1();
//init2();
int _=1;
//cin>>_;
while(_--)solve();
return 0;
}E题
题目大意：
给定两个01串，可以对第一个01串进行操作，将当前的第一个字符移动到字符的末尾，求解A=B的最小操作次数
思路：
首先，我的思路是暴力，可是substr的时间复杂度是O(k)O(k)O(k)，很显然会超时，之后可以考虑到本题使用哈希， 但是数据超过1e4很容易哈希冲突，所以需要使用双哈希
代码：
cpp
#include<bits/stdc++.h>
#define int long long
//#define ull unsigned long long
using namespace std;
//#define x first
//#define y second
//const int mod=19650827;
//const int N=1e5+10;
//int fac[N],invfac[N];
//int qmi(int a,int b,int p){int res=1;while(b){if(b&1)res=res*a%p;a=a*a%p;b>>=1;}return res;}
//int inv(int x){return qmi(x,p-2);}
//int C(int n,int m){if(n<0||m<0||n<m)return 0;return fac[n]*invfac[n-m]%p*invfac[m]%p;}
//void init(){fac[0]=1;for(int i=1;i<N;i++)fac[i]=fac[i-1]*i%p;invfac[N-1]=inv(fac[N-1]);for(int i=N-2;i>=0;i--)invfac[i]=invfac[i+1]*(i+1)%p;}
//struct node{int v,w;bool operator < (const node &x)const{return w==x.w ? v>x.v : w>x.w;}};
//void dijkstra(const vector<vector<node>>&g,vector<int>&dist,vector<bool>&vis,int x){fill(dist.begin(),dist.end(),LLONG_MAX);fill(vis.begin(),vis.end(),false);
//dist[x]=0;priority_queue<node>q;q.push({x,0});while(q.size()){auto u=q.top().v;q.pop();if(vis[u])continue;vis[u]=true;for(auto [v,w]:g[u]){if(dist[v]>dist[u]+w){
//dist[v]=dist[u]+w;q.push({v,dist[v]});}}}}
//struct BIT{int n;vector<int>t;BIT(int _n=0){init(_n);}void init(int _n){n=_n;t.assign(n+2,0);}
//void add(int x,int v){for(;x<=n;x+=(x & -x))t[x]+=v;}int query(int x){int res=0;for(;x;x-=(x & -x))res+=t[x];return res;}};
//int calc_add(const vector<vector<int>>&s,int x1,int y1,int x2,int y2){return s[x2][y2]-s[x2][y1-1]-s[x1-1][y2]+s[x1-1][y1-1];}
//void calc_diff(vector<vector<int>>&d,int x1,int y1,int x2,int y2,int c){d[x1][y1]+=c;d[x2+1][y1]-=c;d[x1][y2+1]-=c;d[x2+1][y2+1]+=c;}
//int primes[N],cnt=0;bool st[N];
//void init1(){for(int i=2;i<N;i++){if(!st[i])primes[++cnt]=i;for(int j=1;primes[j]*i<N;j++){st[primes[j]*i]=true;if(i%primes[j]==0)break;}}}
//vector<int>factor[N];void init2(){for(int i=1;i<N;i++){for(int j=i;j<N;j+=i){factor[j].push_back(i);}}}
//int find(vector<int>&pre,int x){return pre[x]=(pre[x]==x ? x : find(pre,pre[x]));}
//void merge(vector<int>&pre,int x,int y){int fx=find(pre,x),fy=find(pre,y);if(fx==fy)return;pre[fx]=fy;}
//bool check(vector<int>&pre,int x,int y){return find(pre,x)==find(pre,y);}
//string add(string x,string y){if((int)x.size()<(int)y.size())swap(x,y);int n=x.size();int m=y.size();reverse(x.begin(),x.end());reverse(y.begin(),y.end());
//int d=0;string ans;for(int i=0;i<n;i++){int x1=(x[i]-'0');int y1=(i<m ? y[i]-'0' : 0);int sum=x1+y1+d;ans.push_back((sum%10)+'0');d=sum/10;}
//if(d>0)ans.push_back(d+'0');reverse(ans.begin(),ans.end());return ans;}
//string mul(string x,string y){if(x=="0"||y=="0")return "0";reverse(x.begin(),x.end());reverse(y.begin(),y.end());int x1=x.size();int y1=y.size();string ans;
//for(int i=0;i<x1;i++){int d=0;int val1=(x[i]-'0');for(int j=0;j<y1;j++){int val2=(y[j]-'0');int len=ans.size();int val=val1*val2+d;
//if(len<=i+j)ans.push_back((val%10)+'0');else val+=(ans[i+j]-'0'),ans[i+j]=((val%10)+'0');d=val/10;}int len=ans.size();int cur_pos=i+y1;
//while(d){if(cur_pos>=len)ans.push_back((d%10)+'0');else d+=ans[cur_pos]-'0',ans[cur_pos]=((d%10)+'0');d/=10;cur_pos++;}}reverse(ans.begin(),ans.end());return ans;}
//struct Seg_Tree{int n;vector<int>t,lz;Seg_Tree(int _n=0){init(_n);}void init(int _n){n=_n;t.assign((n<<2)+2,0);lz.assign((n<<2)+2,0);}
//void pushup(int o){t[o]=t[o<<1]+t[o<<1|1];}void update(int s,int e,int o,int x){t[o]+=x*(e-s+1);lz[o]+=x;}
//void pushdown(int s,int e,int o){if(!lz[o])return ;int mid=(s+e)>>1;update(s,mid,o<<1,lz[o]);update(mid+1,e,o<<1|1,lz[o]);lz[o]=0;}
//void add(int l,int r,int x,int s=1,int e=0,int o=1){if(e==0)e=this->n;if(l>e||r<s)return ;if(l<=s&&e<=r){update(s,e,o,x);return ;}pushdown(s,e,o);int mid=(s+e)>>1;
//add(l,r,x,s,mid,o<<1);add(l,r,x,mid+1,e,o<<1|1);pushup(o);}
//int query(int l,int r,int s=1,int e=0,int o=1){if(e==0)e=this->n;if(l>e||r<s)return 0;if(l<=s&&r>=e)return t[o];pushdown(s,e,o);int mid=(s+e)>>1;
//int x=query(l,r,s,mid,o<<1);int y=query(l,r,mid+1,e,o<<1|1);return x+y;}};
void solve(){
const int mod1=998244353;
const int mod2=1e9+7;
const int BASE1=2;
const int BASE2=3;
string A,B;
cin>>A>>B;
int len=A.size();
A=A+A;
int n=A.size();
vector<int>hs1(n),hs2(n);
vector<int>pow1(n),pow2(n);
pow1[0]=1;
pow2[0]=1;
for(int i=1;i<n;i++)pow1[i]=pow1[i-1]*BASE1%mod1;
for(int i=1;i<n;i++)pow2[i]=pow2[i-1]*BASE2%mod2;
hs1[0]=A[0]-'0',hs2[0]=A[0]-'0';
for(int i=1;i<n;i++)hs1[i]=(hs1[i-1]*BASE1+(A[i]-'0'))%mod1;
for(int i=1;i<n;i++)hs2[i]=(hs2[i-1]*BASE2+(A[i]-'0'))%mod2;
int hs3=0,hs4=0;
for(int i=0;i<len;i++)hs3=(hs3*BASE1+(B[i]-'0'))%mod1;
for(int i=0;i<len;i++)hs4=(hs4*BASE2+(B[i]-'0'))%mod2;
int ans=-1;
for(int i=0;i<len;i++){
int j=i+len-1;
int val1=((hs1[j]-((i==0 ? 0 : hs1[i-1]*pow1[len])%mod1))+mod1)%mod1;
int val2=((hs2[j]-((i==0 ? 0 : hs2[i-1]*pow2[len])%mod2))+mod2)%mod2;
if(val1==hs3&&val2==hs4){
ans=i;
break;
}
}
cout<<ans<<'\n';
}
signed main(){
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
//init();
//init1();
//init2();
int _=1;
cin>>_;
while(_--)solve();
return 0;
}