【HDU 4913】Least common multiple【线段树】

题意：给出a序列与b序列，set S={x₁,x₂,…,x_n}, where x_i=2^ai * 3^bi.求S集合中所有非空子集合的元素的最小公倍数的和。

思路：对于S序列中的每一个数可以看成是（ai，bi），我们对其按照ai从小到大排序。然后建立一棵以bi序列的线段树。每次加入一个（ai，bi），我们可以知道之前加入的数字的a都比ai小，然后先统计线段树中小于bi的节点个数为cnt，这些数字随意组合子集，此时总共情况为2^cnt,每种情况所产生的最小公倍数为2^ai*3^bi。然后我们统计线段树中大于bi的节点所能产生的最小公倍数情况，所以线段树中我要保留这些情况数×节点所代表的3^bi次方的和。然后利用线段树求区间和sum，sum×2^ai也是这个点所贡献的答案值。计算完之后，将线段树中bi的点加上2^cnt×3^bi的值。这里注意需要离散化。

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define N 100007#define mod 1000000007typedef long long ll;#define lc (d<<1)#define rc (d<<1|1)#define mid (l+r>>1)struct Tr{    int cnt, lz;    ll sum;}tr[N<<4];struct Po{    int x, y;    bool operator<(const Po a) const {        if (x == a.x) return y < a.y;        return x < a.x;    }}po[N];ll cal(ll a, int b) {    ll x = 1;    while (b) {        if (b&1) x = x*a%mod;        a = a*a%mod;        b>>=1;    }    return x;}int b[N];void Push(int d) {    tr[d].sum = tr[lc].sum+tr[rc].sum;    if (tr[d].sum >= mod) tr[d].sum -= mod;    tr[d].cnt = tr[lc].cnt+tr[rc].cnt;}void lazy(int d) {    if (tr[d].lz) {        ll tm = cal(2, tr[d].lz);        tr[lc].sum = tr[lc].sum*tm%mod, tr[rc].sum = tr[rc].sum*tm%mod;        tr[lc].lz += tr[d].lz, tr[rc].lz += tr[d].lz;        tr[d].lz = 0;    }}void build(int d, int l, int r) {    tr[d].lz = 0;    if (l == r) {        tr[d].sum = 0, tr[d].cnt = 0;        return;    }    build(lc, l, mid);    build(rc, mid+1, r);    Push(d);}int C(int d, int l, int r, int L, int R) {    if (L > R) return 0;    if (l == L && r == R) {        return tr[d].cnt;    }    lazy(d);    if (R <=  mid) return C(lc, l, mid, L, R);    else if (L > mid) return C(rc, mid+1, r, L, R);    else return C(lc, l, mid, L, mid)+C(rc, mid+1, r, mid+1, R);}ll S(int d, int l, int r, int L, int R) {    if (L > R) return 0;    if (l == L && r == R) {        return tr[d].sum;    }    lazy(d);    if (R <= mid) return S(lc, l, mid, L, R);    else if (L > mid) return S(rc, mid+1, r, L, R);    else return (S(lc, l, mid, L, mid)+S(rc, mid+1, r, mid+1, R))%mod;}void update(int d, int l, int r, int L, int R) {    if (l == L && r == R) {        tr[d].lz++;        tr[d].sum = tr[d].sum*2%mod;        return;    }    lazy(d);    if (R <= mid) update(lc, l, mid, L, R);    else if (L > mid) update(rc, mid+1, r, L, R);    else update(lc, l, mid, L, mid), update(rc, mid+1, r, mid+1, R);    Push(d);}void add(int d, int l, int r, int pos, ll v) {    if (l == r && l == pos) {        tr[d].sum += v;        if (tr[d].sum >= mod) tr[d].sum -= mod;        tr[d].cnt++;        return;    }    lazy(d);    if (pos <= mid) add(lc, l, mid, pos, v);    else add(rc, mid+1, r, pos, v);    Push(d);}int main(){    int n, i, j;    while (~scanf("%d", &n)) {        for (i = 1;i <= n;i++) {            scanf("%d%d", &po[i].x, &po[i].y);            b[i] = po[i].y;        }        sort(b+1, b+1+n);        int m = unique(b+1, b+1+n)-b-1;        build(1, 1, m);        ll ans = 0;        sort(po+1, po+n+1);        for (i = 1;i <= n;i++) {            int id = lower_bound(b+1, b+1+m, po[i].y)-b;            int cnt = C(1, 1, m, 1, id-1);            ll tm = cal(2, cnt), tp = cal(3, po[i].y), tk = cal(2, po[i].x);            ans = ans+tm*tp%mod*tk%mod;            while (ans >= mod) ans -= mod;            ll sum = S(1, 1, m, id, m);            ans = ans+sum*tk%mod;            while (ans >= mod) ans -= mod;         //   printf("%I64d\n", ans);            update(1, 1, m, id, m);            add(1, 1, m, id, tm*tp%mod);        }        printf("%I64d\n", ans);    }}