【bzoj3813】: 奇数国 数论-线段树-欧拉函数

【bzoj3813】: 奇数国

题意：给定一个序列，每个元素可以分解为最小的60个素数的形式。（x=p1^k1*p2^k2*......p60^k60）（p1=2,p2=3,…,p60=281）

支持单点修改，查询一段区间的积的欧拉函数 mod 19961993（是一个质数）。

线段树维护区间积x，bitset b[i]记录第i个素数是否存在。

预处理inv[i]=(p[i]-1)/p[i] mod 19961993

ans=x*inv[i] (b[i]==1)

/* http://www.cnblogs.com/karl07/ */
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <bitset>
#define LL long long
using namespace std;
const int P=19961993;
const int N=100000;
const int prime[60]={2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181,191,193,197,199,211,223,227,229,233,239,241,251,257,263,269,271,277,281};  
const int inv[60]={9980997,13307996,7984798,11406854,14517814,18426456,9393880,5253157,16490343,8260136,2575742,18343454,3895024,17640832,1698894,3013132,7443456,4581442,9236147,18275065,6562848,2779519,7936697,4037258,6379607,19566707,13566404,4104336,3662752,13602421,16661192,1219054,13259427,9047523,3751248,8196316,14621843,1714528,12192356,11884887,8029406,13455046,17976246,13342473,14084859,15548287,10217514,9846724,5364237,3486812,1627803,14950615,1076789,12406658,19340609,8652728,7791857,7955334,1657495,8808852};  

struct data{ 
    bitset <70> b;
    LL x;
    data (LL x=1){
        if (x==1){
            this->x=1;
            this->b.reset();
        }else{
            this->x=x;
            for (int i=0;i<60;i++){
                if (!(x%prime[i])) this->b[i]=1;
            }
        }
    }
};
data operator + (const data &a,const data &b){ 
    data c; 
    c.b=a.b|b.b; c.x=a.x*b.x%P; 
    return c;
}
struct seg{
    data d;
    seg *ls,*rs;
} t[N*2+5];

seg *root,*NEW=t;
seg *new1(){ return ++NEW;}

#define mid (ll+rr)/2
void build(seg *p,int ll,int rr){
    if (ll==rr){
        p->d=data(3);
    }else{
        build(p->ls=new1(),ll,mid);
        build(p->rs=new1(),mid+1,rr);
        p->d=p->ls->d+p->rs->d;
    }
}

void modify(seg *p,int ll,int rr,int x,data d){
    if (ll==rr){
        p->d=d;
    }else{
        if (x<=mid) modify(p->ls,ll,mid,x,d);
        if (x>mid) modify(p->rs,mid+1,rr,x,d);
        p->d=p->ls->d+p->rs->d;
    }
}

data query(seg *p,int ll,int rr,int l,int r){
    data ans=data(1);
    if (l<=ll && r>=rr){
        ans=ans+p->d;
    }else{
        if (l<=mid) ans=ans+query(p->ls,ll,mid,l,r);
        if (r>mid) ans=ans+query(p->rs,mid+1,rr,l,r);
    }
    return ans;
}

LL cal(data d){
    LL ans=d.x;
    for (int i=0;i<60;i++){
        if (d.b[i]==1){
            ans=ans*inv[i]%P;
        }
    }
    return ans;
}

int n;

int main(){
    scanf("%d",&n);
    build(root=new1(),1,N);
    for (int i=1;i<=n;i++){
        int a,b,c;
        scanf("%d%d%d",&a,&b,&c);
        if (a==0){
            printf("%lld
",cal(query(root,1,N,b,c)));
        }else{
            modify(root,1,N,b,data(c));
        }
    }
    return 0;
}