HDU 4913 Least common multiple 解题报告（线段树）

Least common multiple

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 83    Accepted Submission(s): 25

Problem Description

bobo has an integer set S={x₁,x₂,…,x_n}, where x_i=2^ai * 3^bi.

For each non-empty subsets of S, bobo added the LCM (least common multiple) of the subset up. Find the sum of LCM modulo (10⁹+7).

 

Input

The input consists of several tests. For each tests:

The first line contains n (1≤n≤10⁵). Each of the following n lines contain 2 integers a_i,b_i (0≤a_i,b_i≤10⁹).

 

Output

For each tests:

A single integer, the value of the sum.

 

Sample Input

20 11 031 22 11 2

 

Sample Output

11174

 

Author

Xiaoxu Guo (ftiasch)

 

Source

2014 Multi-University Training Contest 5

    解题报告：思路和官方的思路是一样的。使用线段树维护 b 这边的和。

    官方解题报告路径：blog.sina.com.cn/s/blog_6bddecdc0102uz53.html

    代码如下：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include <queue>#include <vector>#include <map>#include <set>#include <string>#include <iomanip>using namespace std;#define ff(i, n) for(int i=0;i<(n);i++)#define fff(i, n, m) for(int i=(n);i<=(m);i++)#define dff(i, n, m) for(int i=(n);i>=(m);i--)#define bit(n) (1LL<<(n))typedef long long LL;typedef unsigned long long ULL;const LL inf=1e15;void work();int main(){#ifdef ACM    freopen("input.in", "r", stdin);//    freopen("input.in", "w", stdout);#endif // ACM    work();}/***************************************************/const int maxn = 111111;const int mod = 1e9+7;int n;struct Node{    int a, b;    bool operator<(const Node & cmp) const    {        return a == cmp.a ? b < cmp.b : a < cmp.a;    }} node[maxn];double bb[maxn];int tot;int powMod(LL a, int b){    LL res = 1;    while(b)    {        if(b&1)            res = res * a % mod;        a = a * a % mod;        b >>= 1;    }    return res;}#define lson l, m, pos<<1#define rson m+1, r, pos<<1|1#define defm int m = (l+r)/2int sum[maxn<<2];int mul[maxn<<2];void updateFather(int pos){    sum[pos] = (sum[pos<<1] + sum[pos<<1|1]) % mod;}void updateSon(int pos){    if(mul[pos] > 1)    {        mul[pos<<1] = (LL)mul[pos<<1] * mul[pos] % mod;        sum[pos<<1] = (LL)sum[pos<<1] * mul[pos] % mod;        mul[pos<<1|1] = (LL)mul[pos<<1|1] * mul[pos] % mod;        sum[pos<<1|1] = (LL)sum[pos<<1|1] * mul[pos] % mod;        mul[pos] = 1;    }}void updateP(int p, int v, int l, int r, int pos){    if(l == r)    {        sum[pos] = (LL)v * mul[pos] % mod;        return ;    }    updateSon(pos);    defm;    if(p<=m)        updateP(p, v, lson);    else        updateP(p, v, rson);    updateFather(pos);}void update(int L, int R, int l, int r, int pos){    if(L<=l && r<=R)    {        mul[pos] = mul[pos] * 2 % mod;        sum[pos] = sum[pos] * 2 % mod;        return;    }    updateSon(pos);    defm;    if(L<=m)        update(L, R, lson);    if(m<R)        update(L, R, rson);    updateFather(pos);}int query(int L, int R, int l, int r, int pos){    if(L<=l && r<=R)        return sum[pos];    updateSon(pos);    defm;    return ((L<=m?query(L,R,lson):0) + (m<R?query(L,R,rson):0))%mod;}void work(){    while(scanf("%d", &n) == 1)    {        tot = 0;        ff(i, n) scanf("%d%d", &node[i].a, &node[i].b), bb[tot++] = node[i].b;        sort(node, node + n);        sort(bb, bb + tot);        double sub = 0.999999;        double eve = 0.000001;        memset(sum, 0, sizeof(sum));        ff(i, maxn<<2) mul[i] = 1;        LL ans = 0;        ff(i, n)        {            LL now = powMod(2, node[i].a);            int pos = lower_bound(bb, bb+tot, node[i].b) - bb;            bb[pos] -= sub, sub -= eve;            updateP(pos, powMod(3, node[i].b), 0, tot, 1);            ans += now * query(pos, tot, 0, tot, 1) % mod;            update(pos+1, tot, 0, tot, 1);        }        printf("%d\n", ans%mod);    }}