Gcd & Lcm game

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 978    Accepted Submission(s): 217

Problem Description

  Tired of playing too much computer games, alpc23 is planning to play a game on numbers. Because plus and subtraction is too easy for this gay, he wants to do some gcd and lcm operations in a number sequence. After playing it a few times, he has found it is also too boring. So he plan to do a more challenge job: he wants to change several numbers in this sequence and also work out the gcd or lcm of all the number in a subsequence of the whole sequence.
  To be a friend of this gay, you have been invented by him to play this interesting game with him. Of course, you need to work out the answers faster than him to get a free lunch, He he…

 

Input

There are multiple test cases.
For each test case.The first line is the length of sequence n, and the number of queries q. (1<=n, q<=100000) The second line has n numbers, they are the initial n numbers of the sequence a1,a2, …,an, 
From the third line to the q+2 line are the description of the q operations. They are the one of the two forms:
L k1 k2 p; you need to work out the value after mod p of lcm of the subsequence from k1 to k2, inclusive. (1<=k1<=k2<=n) 
G k1 k2 p; you need to work out the value after mod p of gcd of the subsequence from k1 to k2, inclusive. (1<=k1<=k2<=n) 
C k v; the k-th number of the sequence has been changed to v.
You can assume that all the numbers before and after the replacement are positive and no larger than 100.

 

Output

For each of the first or second operation, you need to output the answer in each line.

 

Sample Input

6 41 2 4 5 6 3L 2 5 17G 4 6 4C 4 9G 4 6 4

 

Sample Output

913

ps其实这道题，如果朴素的做法做的话，不难想到，但是由于时间卡的很紧，只能用状态压缩的方法（压缩成int然后用位运算）来加快运算速度 

 T了好几 发，也没想到怎么搞，最后还看大佬博客才会的，很巧妙。 

 除了状态转化外，其他的就是朴素的线段树了。

#include <bits/stdc++.h>using namespace std;typedef long long LL;const int INF = 0x3f3f3f3f;const int maxn = 100000+10;int prime[]={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};int dpos[]={28,25,23,21,20,19,18,17,16,15,14,13,12,11,10,9,8,7,6,5,4,3,2,1,0};int a[]={1,2,4,8,16,32,64};int b[]={1,3,9,27,81};int c[]={1,5,25};int d[]={1,7,49};int num[maxn],N,M;struct node{    int l,r;    int Data_lcm,Data_gcd;}Nodes[maxn<<2];inline int change_input_to_int(int x){    int cnt,y;    y = 0;    for(int i = 0; i < 25 && x>1; i++){        cnt = 0;        while(x%prime[i] == 0){            x/=prime[i];            cnt++;        }        y |= cnt<<dpos[i];    }    return y;}inline int change_int_to_ans(int x,int p){    LL y = 1;    int k;    k = x>>dpos[0];y = y*a[k]%p;x = x^(k<<dpos[0]);    k = x>>dpos[1];y = y*b[k]%p;x = x^(k<<dpos[1]);    k = x>>dpos[2];y = y*c[k]%p;x = x^(k<<dpos[2]);    k = x>>dpos[3];y = y*d[k]%p;x = x^(k<<dpos[3]);    for(int i = 4; i < 25; i++)        if(x  & (1<<dpos[i]))            y = y*prime[i]%p;    return y;}inline int max_lcm(int x,int y){    int ans = 0;    ans |= max(x&0x70000000,y&0x70000000);    ans |= max(x&0x0e000000,y&0x0e000000);    ans |= max(x&0x01800000,y&0x01800000);    ans |= max(x&0x00600000,y&0x00600000);    ans |= ((x&0x001fffff)|(y&0x001fffff));    return ans;}inline int min_gcd(int x,int y){    int ans = 0;    ans |= min(x&0x70000000,y&0x70000000);    ans |= min(x&0x0e000000,y&0x0e000000);    ans |= min(x&0x01800000,y&0x01800000);    ans |= min(x&0x00600000,y&0x00600000);    ans |= ((x&0x001fffff)&(y&0x001fffff));    return ans;}inline void update_father(int now){    Nodes[now].Data_lcm = max_lcm(Nodes[now<<1].Data_lcm,Nodes[now<<1|1].Data_lcm);    Nodes[now].Data_gcd = min_gcd(Nodes[now<<1].Data_gcd,Nodes[now<<1|1].Data_gcd);}void segtree_build(int now,int l,int r){    Nodes[now].l = l;    Nodes[now].r = r;    if(l == r){        Nodes[now].Data_lcm = Nodes[now].Data_gcd = change_input_to_int(num[l]);        return;    }    int mid = (l+r)>>1;    segtree_build(now<<1,l,mid);    segtree_build(now<<1|1,mid+1,r);    update_father(now);}void segtree_update(int now,int pos,int val){    if(Nodes[now].l == Nodes[now].r){        Nodes[now].Data_lcm = Nodes[now].Data_gcd = change_input_to_int(val);        return;    }    int mid = (Nodes[now].l+Nodes[now].r)>>1;    if(pos <= mid)        segtree_update(now<<1,pos,val);    else        segtree_update(now<<1|1,pos,val);    update_father(now);}int segtree_query(int now,int x,int y,char ch){    int res;    if(ch == 'L'){        res = 0;        if(x <= Nodes[now].l && Nodes[now].r <= y){            return Nodes[now].Data_lcm;        }        int mid = (Nodes[now].l+Nodes[now].r)>>1;        if(x <= mid)            res = max_lcm(res,segtree_query(now<<1,x,y,ch));        if(y > mid)            res = max_lcm(res,segtree_query(now<<1|1,x,y,ch));    }    else{        res = 0x7fffffff;        if(x <= Nodes[now].l && Nodes[now].r <= y){            return Nodes[now].Data_gcd;        }        int mid = (Nodes[now].l+Nodes[now].r)>>1;        if(x <= mid)            res = min_gcd(res,segtree_query(now<<1,x,y,ch));        if(y > mid)            res = min_gcd(res,segtree_query(now<<1|1,x,y,ch));    }    return res;}int main(){    char ch[5];    int x,y,id,val,p;    while(~scanf("%d %d",&N,&M)){        for(int i = 1; i <= N; i++)            scanf("%d",&num[i]);        segtree_build(1,1,N);        while(M--){            scanf("%s",ch);            if(ch[0] == 'L' || ch[0] == 'G'){                scanf("%d %d %d",&x,&y,&p);                int result = segtree_query(1,x,y,ch[0]);                printf("%d\n",change_int_to_ans(result,p));            }            else{                scanf("%d %d",&id,&val);                segtree_update(1,id,val);            }        }    }    return 0;}