hdu 3071 Gcd & Lcm game

一道还不错的题目，解法：线段树＋位压缩

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

#include<cstdio>const int maxn=444444;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};#define lson l,mid,lrt#define rson mid+1,r,rrt#define mid ((l+r)>>1)#define lrt rt<<1#define rrt rt<<1|1int MAX[maxn],MIN[maxn];inline int turn(int x){    int cnt,y=0;    for(int i=0;i<25&&x>1;i++){    for(cnt=0;x%prime[i]==0;x/=prime[i]) cnt++;    y|=cnt<<dpos[i];    }    return y;}inline int back(int x,int p){    long long y=1;    int k=x>>dpos[0];y=y*a[k]%p;x^=k<<dpos[0];    k=x>>dpos[1];y=y*b[k]%p;x^=k<<dpos[1];    k=x>>dpos[2];y=y*c[k]%p;x^=k<<dpos[2];    k=x>>dpos[3];y=y*d[k]%p;x^=k<<dpos[3];    for(int i=4;i<25;i++)    if(x&(1<<dpos[i])) y=y*prime[i]%p;    return y;}#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));}inline void pushup(int rt){    MAX[rt]=max(MAX[lrt],MAX[rrt]);    MIN[rt]=min(MIN[lrt],MIN[rrt]);}void build(int l,int r,int rt){    if(l==r){    int x;scanf("%d",&x);    MIN[rt]=MAX[rt]=turn(x);    return;    }    build(lson);build(rson);    pushup(rt);}void update(int k,int x,int l,int r,int rt){    if(l==r){    MAX[rt]=MIN[rt]=turn(x);    return;    }    if(k<=mid) update(k,x,lson);    else update(k,x,rson);    pushup(rt);}int query_max(int s,int t,int l,int r,int rt){    if(s<=l&&t>=r) return MAX[rt];    int ret=0;    if(s<=mid) ret=max(ret,query_max(s,t,lson));    if(t>mid)  ret=max(ret,query_max(s,t,rson));    return ret;}int query_min(int s,int t,int l,int r,int rt){    if(s<=l&&t>=r) return MIN[rt];    int ret=0x7fffffff;    if(s<=mid) ret=min(ret,query_min(s,t,lson));    if(t>mid)  ret=min(ret,query_min(s,t,rson));    return ret;}int main(){    int n,q;    while(scanf("%d%d",&n,&q)!=EOF){    build(1,n,1);    char s[2];    while(q--){        scanf("%s",s);        if(s[0]=='C'){        int k,v;        scanf("%d%d",&k,&v);        update(k,v,1,n,1);        }        else if(s[0]=='L'){        int k1,k2,p;        scanf("%d%d%d",&k1,&k2,&p);        int x=query_max(k1,k2,1,n,1);        printf("%u\n",back(x,p));        }        else{        int k1,k2,p;        scanf("%d%d%d",&k1,&k2,&p);        int x=query_min(k1,k2,1,n,1);        printf("%u\n",back(x,p));        }    }    }    return 0;}