BZOJ 3813 奇数国 线段树+数论

题目大意：给定一个序列，每个数都由60个最小的素数的乘积构成，求某段的乘积的欧拉函数值对19961993取模后的值，支持单点修改

19961993是个质数 出题人还是满贴心的

利用线段树维护乘积取模后的值以及哪些素数出现过 后者用bitset维护

得到的值根据bitset里出现过的素数来计算欧拉函数值

时间复杂度O(nlog10W+60n)

#include <bitset>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define MOD 19961993#define M 100100using namespace std;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 rev[]={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 Segtree{Segtree *ls,*rs;bitset<60>cnt;long long product;void* operator new (size_t){static Segtree mempool[M<<1],*C=mempool;return C++;}void Build_Tree(int x,int y){int mid=x+y>>1;if(x==y){cnt[1]=1;product=3;return ;}ls=new Segtree;rs=new Segtree;ls->Build_Tree(x,mid);rs->Build_Tree(mid+1,y);cnt=ls->cnt|rs->cnt;product=ls->product*rs->product%MOD;}void Modify(int x,int y,int pos,int val){int mid=x+y>>1;if(x==y){int i;for(i=0;i<60;i++)if(val%prime[i]==0)cnt[i]=1;elsecnt[i]=0;product=val;return ;}if(pos<=mid) ls->Modify(x,mid,pos,val);else rs->Modify(mid+1,y,pos,val);cnt=ls->cnt|rs->cnt;product=ls->product*rs->product%MOD;}pair<bitset<60>,long long> Query(int x,int y,int l,int r){int mid=x+y>>1;if(x==l&&y==r)return make_pair(cnt,product);if(r<=mid) return ls->Query(x,mid,l,r);if(l>mid) return rs->Query(mid+1,y,l,r);pair<bitset<60>,long long> temp1=ls->Query(x,mid,l,mid);pair<bitset<60>,long long> temp2=rs->Query(mid+1,y,mid+1,r);return make_pair(temp1.first|temp2.first,temp1.second*temp2.second%MOD);}}*root=new Segtree;int m;int main(){int i,j,p,x,y;cin>>m;root->Build_Tree(1,100000);for(i=1;i<=m;i++){scanf("%d%d%d",&p,&x,&y);if(p==1)root->Modify(1,100000,x,y);else{pair<bitset<60>,long long> temp=root->Query(1,100000,x,y);long long ans=temp.second;for(j=0;j<60;j++)if(temp.first[j])ans*=rev[j],ans%=MOD;printf("%lld\n",ans);}}return 0;}