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:

样例输入:

6 4
1 2 4 5 6 3
L 2 5 17
G 4 6 4
C 4 9
G 4 6 4

样例输出:

9
1
3

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

003 const int maxn=444444;

004 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};

005 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};

006 int a[]={1,2,4,8,16,32,64};

007 int b[]={1,3,9,27,81};

008 int c[]={1,5,25};

009 int 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

017 int MAX[maxn],MIN[maxn];

018 inline 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 }

026 inline 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))

039 inline 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 }

042 inline int max(int x,int y){

044 }

045

046

047 inline void pushup(int rt){

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

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

050 }

051 void 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 }

060 void 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 }

069 int 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 }

076 int 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 }

083 int 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

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

Gcd & Lcm game

题意：

分析：

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