BZOJ3813: 奇数国

传送门

欧拉函数+线段树

因为只有60个素数，所以把状态压成long long的形式。用线段树维护区间和和区间和中有多少个质数。然后xjb搞搞就行了，具体参见代码。

//BZOJ 3813
//by Cydiater
//2016.10.10
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <queue>
#include <map>
#include <ctime>
#include <cstring>
#include <string>
#include <algorithm>
#include <iomanip>
using namespace std;
#define ll long long
#define up(i,j,n)		for(ll i=j;i<=n;i++)
#define down(i,j,n)		for(ll i=j;i>=n;i--)
const ll MAXN=1e6+5;
const ll oo=1LL<<60;
const ll mod=19961993;
const ll LIM=1e6;
const ll N=1e5;
inline ll read(){
	char ch=getchar();ll x=0,f=1;
	while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
ll inv[MAXN],prime[MAXN],cnt=0,T,x,y,k,v;
struct Tree{
	ll S,v;
}t[MAXN<<2];
bool vis[MAXN];
namespace solution{
	void exgcd(ll a,ll b,ll &x,ll &y){
		if(b==0){x=1;y=0;return;}
		exgcd(b,a%b,x,y);
		ll t=x;x=y;y=t-a/b*y;
	}
	ll get_inv(ll a,ll b){
		ll x,y;
		exgcd(a,b,x,y);
		while(x<0)x+=b;
		return x;
	}
	ll get_suf(ll S){
		ll ans=1;
		up(i,0,59)if((1LL<<i)&S)
		ans=(ans*(prime[i+1]-1)%mod*inv[i+1]%mod)%mod;
		return ans;
	}
	ll get_two(ll num){
		ll S=0;
		up(i,1,60)if(num%prime[i]==0)S|=(1LL<<(i-1));
		return S;
	}
	Tree merge(Tree x,Tree y){
		x.S|=y.S;
		(x.v*=y.v)%=mod;
		return x;
	}
	void pret(){
		memset(vis,0,sizeof(vis));
		up(i,2,LIM){
			if(!vis[i])prime[++cnt]=i;
			up(j,1,cnt){
				if(prime[j]*i>LIM)break;
				vis[i*prime[j]]=1;
				if(i%prime[j]==0)break;
			}
		}
		up(i,1,60)inv[i]=get_inv(prime[i],mod);
	}
	void build(int leftt,int rightt,int root){
		if(leftt==rightt){
			t[root].S=1<<1;
			t[root].v=3;
			return;
		}
		int mid=(leftt+rightt)>>1;
		build(leftt,mid,root<<1);
		build(mid+1,rightt,root<<1|1);
		t[root].S=t[root<<1].S|t[root<<1|1].S;
		t[root].v=t[root<<1].v*t[root<<1|1].v%mod;
	}
	Tree get(int leftt,int rightt,int root){
		if(leftt>y||rightt<x)		return (Tree){0,1};
		if(leftt>=x&&rightt<=y)		return t[root];
		int mid=(leftt+rightt)>>1;
		Tree tmp=merge(get(leftt,mid,root<<1),get(mid+1,rightt,root<<1|1));
		return tmp;
	}
	void updata(int leftt,int rightt,int root){
		if(leftt>k||rightt<k)		return;
		if(leftt==rightt){
			t[root].v=v;
			t[root].S=get_two(v);
			return;
		}
		int mid=(leftt+rightt)>>1;
		updata(leftt,mid,root<<1);
		updata(mid+1,rightt,root<<1|1);
		t[root]=merge(t[root<<1],t[root<<1|1]);
	}
	void slove(){
		T=read();
		while(T--){
			ll a=read(),b=read(),c=read();
			if(a==0){
				x=b;y=c;
				Tree tmp=get(1,N,1);
				printf("%lld\n",tmp.v*get_suf(tmp.S)%mod);
			}else{
				k=b;v=c;
				updata(1,N,1);
			}
		}
	}
}
int main(){
	//freopen("input.in","r",stdin);
	using namespace solution;
	pret();
	build(1,N,1);
	slove();
	return 0;
}