【HDOJ】3071 Gcd & Lcm game

刚开始看这个题目，觉得没法做。关键点是数据小于100。因此，可以枚举所有小于100的素因子进行位压缩。
gcd就是求最小值，lcm就是求最大值。c++有时候超时，g++800ms。线段树可解。

/* 3071 */
#include <iostream>
#include <sstream>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#include <deque>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <climits>
#include <cctype>
#include <cassert>
#include <functional>
#include <iterator>
#include <iomanip>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,1024000")

#define sti                set<int>
#define stpii            set<pair<int, int> >
#define mpii            map<int,int>
#define vi                vector<int>
#define pii                pair<int,int>
#define vpii            vector<pair<int,int> >
#define rep(i, a, n)     for (int i=a;i<n;++i)
#define per(i, a, n)     for (int i=n-1;i>=a;--i)
#define clr                clear
#define pb                 push_back
#define mp                 make_pair
#define fir                first
#define sec                second
#define all(x)             (x).begin(),(x).end()
#define SZ(x)             ((int)(x).size())
#define lson            l, mid, rt<<1
#define rson            mid+1, r, rt<<1|1
#define ui32            unsigned int

typedef struct {
    ui32 mx, mn;
} node_t;

int factor[25] = {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 width[25] = {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};
int shift[25] = {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 mask[25] =  {7,7,3,3,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1};
const int maxn = 1e5+5;
int M[25][8];
node_t nd[maxn<<2];
int L, R, mod;

void init() {
    rep(i, 0, 25) {
        M[i][0] = 1;
        rep(j, 1, mask[i]+1)
            M[i][j] = M[i][j-1] * factor[i];
    }
}

int getVal(int x) {
    int ret = 1, tmp;
    
    for (int i=24; i>=0; --i) {
        tmp = M[i][x&mask[i]];
        ret = ret * tmp % mod;
        x >>= width[i];
    }
    
    return ret;
}

ui32 calc(int x) {
    ui32 ret = 0, tmp;
    
    rep(i, 0, 25) {
        ret <<= width[i];
        tmp = 0;
        if (x%factor[i] == 0) {
            while (x%factor[i] == 0) {
                ++tmp;
                x /= factor[i];
            }
        }
        ret += tmp;
    }
    
    return ret;
}

ui32 mymax(ui32 x, ui32 y) {
    ui32 ret = max(x&0x70000000, y&0x70000000)\
            | max(x&0x0e000000, y&0x0e000000)\
            | max(x&0x01800000, y&0x01800000)\
            | max(x&0x00600000, y&0x00600000)\
            | ((x&0x001fffff) | (y&0x001fffff));
    return ret;
}

ui32 mymin(ui32 x, ui32 y) {
    ui32 ret = min(x&0x70000000, y&0x70000000)\
            | min(x&0x0e000000, y&0x0e000000)\
            | min(x&0x01800000, y&0x01800000)\
            | min(x&0x00600000, y&0x00600000)\
            | ((x&0x001fffff) & (y&0x001fffff));
    return ret;
}

inline void PushUp(int rt) {
    int lb = rt << 1;
    int rb = lb |  1;
    
    nd[rt].mx =    mymax(nd[lb].mx, nd[rb].mx);
    nd[rt].mn =    mymin(nd[lb].mn, nd[rb].mn);
}

void Build(int l, int r, int rt) {
    if (l == r) {
        int x;
        scanf("%d", &x);
        nd[rt].mx = nd[rt].mn = calc(x);
        return ;
    }
    
    int mid = (l + r) >> 1;
    
    Build(lson);
    Build(rson);
    PushUp(rt);
}

ui32 Query_mx(int l, int r, int rt) {
    if (L<=l && R>=r) {
        return nd[rt].mx;
    }
    
    int mid = (l + r) >> 1;
    ui32 ret;
    
    if (R <= mid) {
        ret = Query_mx(lson);
    } else if (L > mid) {
        ret = Query_mx(rson);
    } else {
        ui32 ltmp = Query_mx(lson);
        ui32 rtmp = Query_mx(rson);
        ret = mymax(ltmp, rtmp);
    }
    
    return ret;
}

ui32 Query_mn(int l, int r, int rt) {
    if (L<=l && R>=r) {
        return nd[rt].mn;
    }
    
    int mid = (l + r) >> 1;
    ui32 ret;
    
    if (R <= mid) {
        ret = Query_mn(lson);
    } else if (L > mid) {
        ret = Query_mn(rson);
    } else {
        ui32 ltmp = Query_mn(lson);
        ui32 rtmp = Query_mn(rson);
        ret = mymin(ltmp, rtmp);
    }
    
    return ret;
}

void Update(int k, int x, int l, int r, int rt) {
    if (l == r) {
        nd[rt].mx = nd[rt].mn = calc(x);
        return ;
    }
    
    int mid = (l + r) >> 1;
    
    if (k <= mid)
        Update(k, x, lson);
    else
        Update(k, x, rson);
    
    PushUp(rt);
}

int main() {
    ios::sync_with_stdio(false);
    #ifndef ONLINE_JUDGE
        freopen("data.in", "r", stdin);
        freopen("data.out", "w", stdout);
    #endif
    
    int n, q;
    char op[10];
    int tmp, k;
    int ans;
    
    init();
    while (scanf("%d %d", &n, &q) != EOF) {
        Build(1, n, 1);
        while (q--) {
            scanf("%s", op);
            if (op[0] == 'L') {
                scanf("%d %d %d", &L, &R, &mod);
                tmp = Query_mx(1, n, 1);
                ans = getVal(tmp);
                printf("%d\n", ans);
            } else if (op[0] == 'G') {
                scanf("%d %d %d", &L, &R, &mod);
                tmp = Query_mn(1, n, 1);
                ans = getVal(tmp);
                printf("%d\n", ans);
            } else {
                scanf("%d %d", &k, &tmp);
                Update(k, tmp, 1, n, 1);
            }
        }
    }
    
    #ifndef ONLINE_JUDGE
        printf("time = %d.\n", (int)clock());
    #endif
    
    return 0;
}