HDU 3071-Gcd & Lcm game-线段树+素因子分解-[解题报告]HOJ

Gcd & Lcm game

问题描述 :

  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…

输入:

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.

输出:

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.

样例输入:

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

样例输出:

913

一道还不错的题目，解法：线段树＋位压缩

题意：

给定一个长度为n的序列m次操作，操作的种类一共有三种

• 查询 
  
  • L :查询一个区间的所有的数的最小公倍数modp
  • G :查询一个区间的所有的数的最大公约数modp
• 修改 
  
  • C :将给定位置的值修改成x

分析：

首先我们注意一下数据的范围，保证数据不超过100，那么很明显素因子特别少一共只有25个，我们可以用线段树维护一下对应素因子的最大值与最小值。更新的话就是单点更新。由于时间比较紧，我们需要把所有的数压到一个int中去。

对任意x<=100 其因子个数情况如下：

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};//max num          7  4  2  2  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1//bit              3  3  2  2  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  1//tot bit  3+3+2+2+21*1=31// 0000 0000 0000 0000 0000 0000 0000 0000//    |   |  | |                             //    2   3  5 7 

所以，可以用一个32位的int数字表示x对应的各因子数

#define _min(x,y) ((x)<(y)?(x):(y))#define _max(x,y) ((x)>(y)?(x):(y))inline int min(int x,int y){    return _min(x&0x70000000,y&0x70000000)|_min(x&0x0e000000,y&0x0e000000)|_min(x&0x01800000,y&0x01800000)|_min(x&0x00600000,y&0x00600000)|((x&0x001fffff)&(y&0x001fffff));}inline int max(int x,int y){    return _max(x&0x70000000,y&0x70000000)|_max(x&0x0e000000,y&0x0e000000)|_max(x&0x01800000,y&0x01800000)|_max(x&0x00600000,y&0x00600000)|((x&0x001fffff)|(y&0x001fffff));}

自定义比较函数，即分别计算x与y的每个因子出现的最大与最小次数。

然后就是裸的单点更新，成段求最值的线段树了

001#include<cstdio>

002 

003const int maxn=444444;

004int 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};

005int  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};

006int a[]={1,2,4,8,16,32,64};

007int b[]={1,3,9,27,81};

008int c[]={1,5,25};

009int d[]={1,7,49};

010 

011#define lson l,mid,lrt

012#define rson mid+1,r,rrt

013#define mid ((l+r)>>1)

014#define lrt rt<<1

015#define rrt rt<<1|1

016 

017int MAX[maxn],MIN[maxn];

018inline int turn(int x){

019    int cnt,y=0;

020    for(int i=0;i<25&&x>1;i++){

021    for(cnt=0;x%prime[i]==0;x/=prime[i]) cnt++;

022    y|=cnt<<dpos[i];

023    }

024    return y;

025}

026inline int back(int x,int p)

027{

028    long long y=1;

029    int k=x>>dpos[0];y=y*a[k]%p;x^=k<<dpos[0];

030    k=x>>dpos[1];y=y*b[k]%p;x^=k<<dpos[1];

031    k=x>>dpos[2];y=y*c[k]%p;x^=k<<dpos[2];

032    k=x>>dpos[3];y=y*d[k]%p;x^=k<<dpos[3];

033    for(int i=4;i<25;i++)

034    if(x&(1<<dpos[i])) y=y*prime[i]%p;

035    return y;

036}

037#define _min(x,y) ((x)<(y)?(x):(y))

038#define _max(x,y) ((x)>(y)?(x):(y))

039inline int min(int x,int y){

040    return _min(x&0x70000000,y&0x70000000)|_min(x&0x0e000000,y&0x0e000000)|_min(x&0x01800000,y&0x01800000)|_min(x&0x00600000,y&0x00600000)|((x&0x001fffff)&(y&0x001fffff));

041}

042inline int max(int x,int y){

043    return _max(x&0x70000000,y&0x70000000)|_max(x&0x0e000000,y&0x0e000000)|_max(x&0x01800000,y&0x01800000)|_max(x&0x00600000,y&0x00600000)|((x&0x001fffff)|(y&0x001fffff));

044}

045 

046 

047inline void pushup(int rt){

048    MAX[rt]=max(MAX[lrt],MAX[rrt]);

049    MIN[rt]=min(MIN[lrt],MIN[rrt]);

050}

051void build(int l,int r,int rt){

052    if(l==r){

053    int x;scanf("%d",&x);

054    MIN[rt]=MAX[rt]=turn(x);

055    return;

056    }

057    build(lson);build(rson);

058    pushup(rt);

059}

060void update(int k,int x,int l,int r,int rt){

061    if(l==r){

062    MAX[rt]=MIN[rt]=turn(x);

063    return;

064    }

065    if(k<=mid) update(k,x,lson);

066    else update(k,x,rson);

067    pushup(rt);

068}

069int query_max(int s,int t,int l,int r,int rt){

070    if(s<=l&&t>=r) return MAX[rt];

071    int ret=0;

072    if(s<=mid) ret=max(ret,query_max(s,t,lson));

073    if(t>mid)  ret=max(ret,query_max(s,t,rson));

074    return ret;

075}

076int query_min(int s,int t,int l,int r,int rt){

077    if(s<=l&&t>=r) return MIN[rt];

078    int ret=0x7fffffff;

079    if(s<=mid) ret=min(ret,query_min(s,t,lson));

080    if(t>mid)  ret=min(ret,query_min(s,t,rson));

081    return ret;

082}

083int main()

084{

085    int n,q;

086    while(scanf("%d%d",&n,&q)!=EOF){

087    build(1,n,1);

088    char s[2];

089    while(q--){

090        scanf("%s",s);

091        if(s[0]=='C'){

092        int k,v;

093        scanf("%d%d",&k,&v);

094        update(k,v,1,n,1);

095        }

096        else if(s[0]=='L'){

097        int k1,k2,p;

098        scanf("%d%d%d",&k1,&k2,&p);

099        int x=query_max(k1,k2,1,n,1);

100        printf("%u\n",back(x,p));

101        }

102        else{

103        int k1,k2,p;

104        scanf("%d%d%d",&k1,&k2,&p);

105        int x=query_min(k1,k2,1,n,1);

106        printf("%u\n",back(x,p));

107        }

108    }

109    }

110    return 0;

111}

参考：http://blog.csdn.net/wxfwxf328/article/details/7479874