莫队大法好

无修改的离线区间问题处理(乱搞)神器——莫队

洛谷P3901 数列找不同

#include<bits/stdc++.h>
#define IL inline
#define R register int

using namespace std;
const int N=1e5+5,inf=0x3f3f3f3f;

IL int read() {
    int f=1;
    char ch;
    while((ch=getchar())<'0'||ch>'9') if(ch=='-') f=-1;
    int res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9') res=res*10+ch-'0';
    return res*f;
}

int n,m,len,cnt[N],curl,curr,ans,a[N];//curl=1和curl=0皆可
bool o[N];
struct hh {
    int l,r,id;
    bool operator <(const hh &a) const {
        if(l/len==a.l/len) return r<a.r;
        return l<a.l;
    }
} q[N];

void add(int x) {
    if((++cnt[a[x]])==1) ans++;
}

void del(int x) {
    if(--(cnt[a[x]])==0) ans--;
}

int main() {
    n=read();
    m=read();
    len=sqrt(n);
    for(R i=1; i<=n; ++i) a[i]=read();
    for(R i=1; i<=m; ++i) q[i].l=read(),q[i].r=read(),q[i].id=i,o[i]=0;
    sort(q+1,q+1+m);

    for(R i=1; i<=m; ++i) {
        while(q[i].l>curl) del(curl++);
        while(q[i].l<curl) add(--curl);
        while(q[i].r>curr) add(++curr);
        while(q[i].r<curr) del(curr--);
        if(ans==(q[i].r-q[i].l+1)) o[q[i].id]=1;
    }
    for(R i=1; i<=m; ++i)
        if(o[i]) puts("Yes");
        else puts("No");
    return 0;
}

洛谷P2709 小B的询问

#include<bits/stdc++.h>
#define int long long
#define IL inline
#define R register int

using namespace std;
const int N=1e5+5,inf=0x3f3f3f3f;

IL int read() {
    int f=1;
    char ch;
    while((ch=getchar())<'0'||ch>'9') if(ch=='-') f=-1;
    int res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9') res=res*10+ch-'0';
    return res*f;
}

int n,m,len,cnt[N],curl=1,curr,ans,a[N],k;//注意！curl=1
int o[N];
struct hh {
    int l,r,id;
    bool operator <(const hh &a) const {
        if(l/len==a.l/len) return r<a.r;
        return l<a.l;
    }
} q[N];

void add(int x) {
    ans+=2*cnt[a[x]]+1;
    cnt[a[x]]++;
}

void del(int x) {
    ans-=2*cnt[a[x]]-1;//完全平方展开
    cnt[a[x]]--;
}

signed main() {
    n=read();
    m=read();
    k=read();
    len=sqrt(n);
    for(R i=1; i<=n; ++i) a[i]=read();
    for(R i=1; i<=m; ++i) q[i].l=read(),q[i].r=read(),q[i].id=i;
    sort(q+1,q+1+m);
    for(R i=1; i<=m; ++i) {
        while(curl<q[i].l) del(curl++);
        while(curl>q[i].l) add(--curl);
        while(curr<q[i].r) add(++curr);
        while(curr>q[i].r) del(curr--);
        o[q[i].id]=ans;
    }
    for(R i=1; i<=m; ++i) printf("%lld
",o[i]);
    return 0;
}

洛谷P4462 [CQOI2018]异或序列

#include<bits/stdc++.h>
#define int long long
#define IL inline
#define R register int

using namespace std;
const int N=2e5+5,inf=0x3f3f3f3f;

IL int read() {
    int f=1;
    char ch;
    while((ch=getchar())<'0'||ch>'9') if(ch=='-') f=-1;
    int res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9') res=res*10+ch-'0';
    return res*f;
}

int n,m,len,cnt[N],curl=1,curr,ans,a[N],k;
int o[N];
struct hh {
    int l,r,id;
    bool operator <(const hh &a) const {
        if(l/len==a.l/len) return r<a.r;
        return l<a.l;
    }
} q[N];

void add(int x) {
    ans+=cnt[a[x]^k];
    cnt[a[x]]++;//顺序Attention!
}

void del(int x) {
    cnt[a[x]]--;
    ans-=cnt[a[x]^k];
}

//区间：[l-1,r]
signed main() {
    n=read();
    m=read();
    k=read();
    len=sqrt(n);
    for(R i=1; i<=n; ++i) a[i]=read(),a[i]^=a[i-1];
    for(R i=1; i<=m; ++i) q[i].l=read()-1,q[i].r=read(),q[i].id=i;//Attention!q[i].l--;
    sort(q+1,q+1+m);
    for(R i=1; i<=m; ++i) {
        while(curl<q[i].l) del(curl++);
        while(curl>q[i].l) add(--curl);
        while(curr<q[i].r) add(++curr);
        while(curr>q[i].r) del(curr--);
        o[q[i].id]=ans;
    }
    for(R i=1; i<=m; ++i) printf("%lld
",o[i]);
    return 0;
}

CF617E XOR and Favorite Number （同上一题，数据范围有所变化）

洛谷P1972 [SDOI2009]HH的项链（著名卡莫队题，然鹅我还是用莫队卡过了，感谢讨论帖的好心人！）

#include<bits/stdc++.h>
//#define int long long
#define IL inline
#define R register int

using namespace std;
const int N=1e6+5,inf=0x3f3f3f3f;

IL int read() {
    int f=1;
    char ch;
    while((ch=getchar())<'0'||ch>'9') if(ch=='-') f=-1;
    int res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9') res=res*10+ch-'0';
    return res*f;
}

int n,m,len,cnt[N],curl=1,curr,ans,a[N],k,b[N];
int o[N];
struct hh {
    int l,r,id;
    bool operator <(const hh &a) const {
        if(b[l]==b[a.l]) 
            if(b[l]&1) return r<a.r;
            else return r>a.r;
        return l<a.l;
    }//奇偶性排序
} q[N];

void add(int x) {
    cnt[a[x]]++;
    ans+=cnt[a[x]]==1;//少用if也会变快
}

void del(int x) {
    cnt[a[x]]--;
    ans-=cnt[a[x]]==0;
}

int main() {
    n=read();
    len=sqrt(n);
    for(R i=1; i<=n; ++i) a[i]=read(),b[i]=(i-1)/len+1;//预处理b数组会更快
    m=read();
    for(R i=1; i<=m; ++i) q[i].l=read(),q[i].r=read(),q[i].id=i;
    sort(q+1,q+1+m);
    for(R i=1; i<=m; ++i) {
        while(curl<q[i].l) del(curl++);
        while(curl>q[i].l) add(--curl);
        while(curr<q[i].r) add(++curr);
        while(curr>q[i].r) del(curr--);
        o[q[i].id]=ans;
    }
    for(R i=1; i<=m; ++i) printf("%d
",o[i]);
    return 0;
}

SP3267 DQUERY - D-query（上一题的弱化版，然鹅至今未过，一直在waiting）

洛谷P1494 [国家集训队]小Z的袜子

#include<bits/stdc++.h>
#define int long long
#define IL inline
#define R register int

using namespace std;
const int N=1e5+5,inf=0x3f3f3f3f;

IL int read() {
    int f=1;
    char ch;
    while((ch=getchar())<'0'||ch>'9') if(ch=='-') f=-1;
    int res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9') res=res*10+ch-'0';
    return res*f;
}

int n,m,len,cnt[N],curl,curr,ans,a[N],k;
int gcd(int x,int y) {return y?gcd(y,x%y):x;}
struct Ans {
    int x,y;
    void out() {
        if(!x) puts("0/1");
        else {
            int d=gcd(x,y);
            x/=d;y/=d;
            printf("%lld/%lld
",x,y);
        }
    }
}o[N];
struct hh {
    int l,r,id;
    bool operator <(const hh &a) const {
        if(l/len==a.l/len) return r<a.r;
        return l<a.l;
    }
} q[N];

void add(int i) {
    if(cnt[a[i]]>0) ans+=cnt[a[i]]*(cnt[a[i]]+1)-cnt[a[i]]*(cnt[a[i]]-1);//也可以这样写:ans+=cnt[a[i]]*2;
    cnt[a[i]]++;
}

void del(int i) {
    if(cnt[a[i]]>1) ans+=(cnt[a[i]]-1)*(cnt[a[i]]-2)-cnt[a[i]]*(cnt[a[i]]-1);//也可以这样写:ans-=(cnt[a[i]]-1)*2;
    cnt[a[i]]--;
}

signed main() {
    n=read();
    m=read();
    len=sqrt(n);
    for(R i=1; i<=n; ++i) a[i]=read();
    for(R i=1; i<=m; ++i) q[i].l=read(),q[i].r=read(),q[i].id=i;
    sort(q+1,q+1+m);
    for(R i=1; i<=m; ++i) {
        while(curl<q[i].l) del(curl++);
        while(curl>q[i].l) add(--curl);
        while(curr<q[i].r) add(++curr);
        while(curr>q[i].r) del(curr--);
        o[q[i].id]=(Ans){ans,(q[i].r-q[i].l+1)*(q[i].r-q[i].l)};
    }
    for(R i=1; i<=m; ++i) o[i].out();
    return 0;
}