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可以参考线段树或分块维护区间加和区间减的思路

看一下数据范围,每次操作只能O(1)的复杂度。

点的位置可以离散化一下

记得先乘后加

#include<bits/stdc++.h>
#include<unordered_map>
namespace ZDY{
    #pragma GCC optimize(3)
    #define il __inline__ __attribute__ ((always_inline))
    #define rg register
    #define ll long long
    #define ull unsigned long long
    #define db double
    #define sht short
    #define MB template <class T>il
    #define Fur(i,x,y) for(int i=x;i<=y;++i)
    #define Fdr(i,x,y) for(int i=x;i>=y;--i)
    #define fl(i,x) for(int i=head[x],to;to=e[i].to,i;i=e[i].nxt)
    #define clr(x,y) memset(x,y,sizeof(x))
    #define cpy(x,y) memcpy(x,y,sizeof(x))
    #define fin(s) freopen(s".in","r",stdin)
    #define fout(s) freopen(s".out","w",stdout)
    #define fcin ios::sync_with_stdio(false)
    #define l2(n) (int(log2(n)))
    #define inf 0x3f3f3f3f
    MB T ABS(T x){return x>0?x:-x;}
    MB T MAX(T x,T y){return x>y?x:y;}
    MB T MIN(T x,T y){return x<y?x:y;}
    MB T GCD(T x,T y){return y?GCD(y,x%y):x;}
    MB void SWAP(T &x,T &y){T t=x;x=y;y=t;}
}using namespace ZDY;using namespace std;
namespace IO{const char* ln="\n";const int str=1<<20;struct IN{char buf[str],*s,*t;bool _;IN():s(buf),t(buf),_(0){}il char gc(){return s==t&&((t=(s=buf)+fread(buf,1,str,stdin))==s)?EOF:(*s++);}IN&operator>>(char&ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)_=1;else ch=c;return *this;}IN& operator>>(char* ch){clr(ch,0);if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)_=1;return *this;}IN& operator>>(string& ch){if(_)return *this;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)return _=1,*this;ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)_=1;return *this;}template<typename T>IN&operator>>(T&x){if(_)return *this;char c=gc();bool ff=0;while(c!=EOF&&(c<'0'||c>'9'))ff^=(c=='-'),c=gc();if(c==EOF){_=1;return *this;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=(x<<3)+(x<<1)+c-48,c=gc();if(c==EOF)_=1;if(ff)x=-x;return *this;}}in;struct OUT{char buf[str],*s,*t;OUT():s(buf),t(buf+str){}~OUT(){fwrite(buf,1,s-buf,stdout);}void pt(char c){(s==t)?(fwrite(s=buf,1,str,stdout),*s++=c):(*s++=c);}OUT&operator<<(const char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(char*s){while(*s)pt(*s++);return *this;}OUT&operator<<(string s){for(int i=0;s[i];i++)pt(s[i]);return *this;}template<typename T>OUT&operator<<(T x){if(!x)return pt('0'),*this;if(x<0)pt('-'),x=-x;char a[30],t=0;while(x)a[t++]=x%10,x/=10;while(t--)pt(a[t]+'0');return *this;}}out;}using namespace IO;
#define N 100007
const int mod=10000019;
int n,Q,add=0,mul=1,ans=0,s=0,v[N],ti[N],la,inv[mod+1];
struct que{
    int u,x,y;
}q[N];
void work(int i,int k){
    int u=q[i].u,x=q[i].x,y=q[i].y;
    if(u==6)ans=(ans+s)%mod;
    else if(u==2)s=(s+1ll*n*x%mod)%mod,add=(add+x)%mod;
    else if(u==3)s=1ll*s*x%mod,add=1ll*add*x%mod,mul=1ll*mul*x%mod;
    else if(u==4)s=1ll*n*x%mod,add=0,mul=1,v[0]=x,la=k;
    else if(u==5){
        if(ti[x]<=la)ans=(ans+1ll*mul*v[0]%mod+add)%mod;
        else ans=(ans+1ll*mul*v[x]%mod+add)%mod;
    }
    else{
        if(ti[x]<=la)s=(s-1ll*mul*v[0]%mod-add+(mod<<1))%mod;
        else s=(s-1ll*mul*v[x]%mod-add+(mod<<1))%mod;
        s=(s+y)%mod;ti[x]=k;v[x]=1ll*(y-add+mod)*inv[mul]%mod;
    }
}
unordered_map<int,int>mp;
int a[N],b[N];
int main(){
    in>>n>>Q;
    inv[1]=1;
    Fur(i,2,mod)inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
    int u,x,d=0;
    Fur(i,1,Q){
        in>>q[i].u;
        if(q[i].u!=6)in>>q[i].x;
        if(q[i].u==1){
            in>>q[i].y,q[i].y=(q[i].y%mod+mod)%mod;
            if(!mp[q[i].x])mp[q[i].x]=++d;
        }
        else if(q[i].u<=4)q[i].x=(q[i].x%mod+mod)%mod;
        if(q[i].u==3&&q[i].x==0)q[i].u=4;
    }
    Fur(i,1,Q)if(q[i].u==1||q[i].u==5)q[i].x=mp[q[i].x];
    mul=1;
    int t;in>>t;
    Fur(i,1,t)in>>a[i]>>b[i];
    Fur(i,1,t*Q){
        int x=(i-1)/Q+1,y=(i-1)%Q+1;
        work((a[x]+1ll*y*b[x])%Q+1,i);
    }
    out<<ans<<ln;
}
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