HDU 4913 Least common multiple

/*
hdu4913 Least common multiple
http://acm.hdu.edu.cn/showproblem.php?pid=4913
离散化 线段树 统计逆序数思想
tips:
1、线段树中一定要到处都取模，否则wa。。。
2、lazy是乘积的形式出现，不是加和
 */
#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
const long long Nmax=100005LL;
const long long mod=1000000007LL;
long long ans;
int n;
vector<long long> v;
struct Node
{
    long long a;
    long long b;
};
Node num[Nmax];
long long a[Nmax],b[Nmax],ab[Nmax];
bool cmp(Node a,Node b)
{
    if(a.a==b.a)
        return a.b<b.b;
    return  a.a<b.a;
}
int get_id(long long x)
{
    return lower_bound(v.begin(),v.end(),x)-v.begin()+1;
}
struct Tree
{
    int l;
    int r;
    long long sum;
    long long times;
    long long lazy;
}tree[Nmax*4];
void build(int root,int l,int r)
{
    tree[root].l=l;
    tree[root].r=r;
    tree[root].sum=tree[root].times=0LL;
    tree[root].lazy=1LL;
    if(l==r)
        return;
    int mid=(l+r)>>1;
    build(root<<1,l,mid);
    build(root<<1|1,mid+1,r);
}

void push_down(int root)
{
    if(tree[root].l==tree[root].r)
    {
        tree[root].lazy=1LL;
        return;
    }
    if(tree[root].lazy!=1LL)
    {
        tree[root<<1].lazy=(tree[root<<1].lazy*tree[root].lazy)%mod;
        tree[root<<1|1].lazy=(tree[root<<1|1].lazy*tree[root].lazy)%mod;
        tree[root<<1].sum=(tree[root<<1].sum*tree[root].lazy)%mod;
        tree[root<<1|1].sum=(tree[root<<1|1].sum*tree[root].lazy)%mod;
        tree[root].lazy=1LL;
    }
}

void insert(int root,int l,int r,int data)
{
    if(tree[root].l==l&&tree[root].r==r)
    {
        tree[root].times+=data;
        return;
    }
    push_down(root);
    int mid=(tree[root].l+tree[root].r)>>1;
    if(mid>=l)
        insert(root<<1,l,r,data);
    if(mid<r)
        insert(root<<1|1,l,r,data);
    tree[root].times=tree[root<<1].times+tree[root<<1|1].times;
}
void insert_sum(int root,int l,int r,int data)
{
    if(tree[root].l>=l&&tree[root].r<=r)
    {
        tree[root].sum=(tree[root].sum+data)%mod;
        return;
    }
    push_down(root);
    int mid=(tree[root].l+tree[root].r)>>1;
    if(mid>=l)
        insert_sum(root<<1,l,r,data);
    if(mid<r)
        insert_sum(root<<1|1,l,r,data);
    tree[root].sum=(tree[root<<1].sum+tree[root<<1|1].sum)%mod;
}

long long query_sum(int root,int l,int r)
{
    if(tree[root].l>=l&&tree[root].r<=r)
    {
        return tree[root].sum%mod;
    }
    push_down(root);
    int mid=(tree[root].l+tree[root].r)>>1;
    long long ans=0LL;
    if(mid>=l)
        ans=( ans+query_sum(root<<1,l,r) )%mod;
    if(mid<r)
        ans=( ans+query_sum(root<<1|1,l,r) )%mod;
    while(ans<0)
        ans+=mod;
    return ans;
}

long long query(int root,int l,int r)
{
    if(tree[root].l>=l&&tree[root].r<=r)
    {
        return tree[root].times;
    }
    int mid=(tree[root].l+tree[root].r)>>1;
    long long ans=0LL;
    if(mid>=l)
        ans+=query(root<<1,l,r);
    if(mid<r)
        ans+=query(root<<1|1,l,r);
    return ans;
}


void init()
{
    v.clear();
    ans=0LL;
    build(1,1,n);
}

int qpow(long long base,long long n)
{
    base=base%mod;
    long long ans=1LL;
    while(n>0)
    {
        if(n&1)
            ans=(ans*base)%mod;
        base=base*base%mod;
        n>>=1;
    }
    while(ans<0)
        ans+=mod;
    return ans;
}

void update(int root,int l,int r)
{
    if(tree[root].l>=l && tree[root].r<=r)
    {
        tree[root].lazy=(tree[root].lazy*2LL)%mod;
        tree[root].sum=(tree[root].sum*2LL)%mod;
        return;
    }
    push_down(root);
    int mid=(tree[root].l+tree[root].r)>>1;
    if(mid>=l)
        update(root<<1,l,r);
    if(mid<r)
        update(root<<1|1,l,r);
    tree[root].sum=(tree[root<<1].sum+tree[root<<1|1].sum)%mod;
}

void watch(int root,int l,int r)
{
    printf("tree[%d]: l:%d,r:%d,sum:%lld,times:%lld,lazy:%lld\n",root,tree[root].l,tree[root].r,tree[root].sum,tree[root].times,tree[root].lazy);
    if(l==r)
        return;
    int mid=(l+r)>>1;
    watch(root<<1,1,mid);
    watch(root<<1|1,mid+1,r);
}
int main()
{
    //freopen("hdu4913.in","r",stdin);
    while(scanf("%d",&n)==1)
    {
        init();
        for(int i=1;i<=n;i++)
        {
            scanf("%lld%lld",&num[i].a,&num[i].b);
            v.push_back(num[i].b);
        }
        sort(v.begin(),v.end());
        v.erase(unique(v.begin(),v.end()),v.end());
        sort(num+1,num+1+n,cmp);
        int m=v.size();
        for(int i=1;i<=n;i++)
        {
                //printf("times[%d]:\n",i);
            int b=get_id(num[i].b);
                //printf("b:%d\n",b);
                //printf("before:\n");
                //watch(1,1,n);
                //printf("after:\n");
                //watch(1,1,n);
                //printf("end!!!!!!!!!!!!!!!!!\n");
            insert(1,b,b,1);
            long long x=query(1,1,b);
                //printf("%lld:%lld\n",query(1,1,b),query(1,b,b)); 
                //printf("x:%lld\n",x); 
            x=qpow(2,x-1);
                //printf("x:%lld\n",x);

            x= (x*qpow(3,num[i].b)) %mod;
                //printf("x:%lld\n",x);
            ans=( ans+ qpow(2,num[i].a)*x )%mod;
                //printf("ans:%lld\n",ans);
            //if(kkk==0)
            //if(b!=n)
            if(b!=n)
                ans=(ans+(qpow(2,num[i].a)* query_sum(1,b+1,n)) %mod)%mod;
                //printf("ans:%lld\n",ans);
            if(b!=n)
                update(1,b+1,n);
            insert_sum(1,b,b,x);
                //watch(1,1,n);
        }
        while(ans<0)
            ans+=mod;
                //printf("$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$\n\n\n");
        printf("%lld\n",ans);
    }
    return 0;
}