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CodeForces 689E Mike and Geometry Problem

时间:2016-09-16 10:15:36      阅读:154      评论:0      收藏:0      [点我收藏+]

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#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<iostream>
using namespace std;
typedef long long LL;
const double pi=acos(-1.0),eps=1e-6;
void File()
{
    freopen("D:\\in.txt","r",stdin);
    freopen("D:\\out.txt","w",stdout);
}
template <class T>
inline void read(T &x)
{
    char c=getchar(); x=0;
    while(!isdigit(c)) c=getchar();
    while(isdigit(c)) {x=x*10+c-0; c=getchar();}
}

LL mod=1e9+7;
const int maxn=1200010;
int n,k;
LL f[maxn],a[maxn];
struct X { int L,R; }s[maxn];
int c[maxn],b[maxn],sz;
LL ans[maxn];

int lowbit(int x){return x&(-x);}
int sum(int x)
{
    int res=0;
    while(x>0) res=res+c[x],x=x-lowbit(x);
    return res;
}
void update(int x,int v)
{
    while(x<=1200000) c[x]=c[x]+v,x=x+lowbit(x);
}

LL extend_gcd(LL a,LL b,LL &x,LL &y)
{
    if(a==0&&b==0) return -1;
    if(b==0){x=1;y=0;return a;}
    LL d=extend_gcd(b,a%b,y,x);
    y-=a/b*x;
    return d;
}

LL mod_reverse(LL a,LL n)
{
    LL x,y;
    LL d=extend_gcd(a,n,x,y);
    if(d==1) return (x%n+n)%n;
    else return -1;
}

int get(int x)
{
    int L=0,R=sz-1,pos=0;
    while(L<=R)
    {
        int mid=(L+R)/2;
        if(b[mid]<x) L=mid+1;
        else if(b[mid]==x) pos=mid,R=mid-1;
        else R=mid-1;
    }
    return pos+1;
}

bool cmp(X a,X b) { return a.L<b.L; }

int main()
{
    scanf("%d%d",&n,&k);
    f[0]=1; for(int i=1;i<=400000;i++) f[i]=i*f[i-1]%mod;
    for(int i=k;i<=400000;i++)
    {
        LL fz=f[i]%mod,fm=f[k]*f[i-k]%mod;
        LL ni=mod_reverse(fm,mod);
        a[i]=fz*ni%mod;
    }

    for(int i=1;i<=n;i++)
    {
        scanf("%d%d",&s[i].L,&s[i].R);
        b[sz++]=s[i].L, b[sz++]=s[i].R;
        b[sz++]=s[i].L-1; b[sz++]=s[i].L+1;
        b[sz++]=s[i].R-1; b[sz++]=s[i].R+1;
    }

    sort(b,b+sz);
    sort(s+1,s+1+n,cmp);

    int h=1;
    for(int i=1;i<=sz;i++)
    {
        while(h<=n&&get(s[h].L)==i)
        {
            update(get(s[h].R),1);
            h++;
        }
        ans[i]=a[sum(1200000)-sum(i-1)];
    }

    LL Ans=0;
    for(int i=0;i<sz;)
    {
        int pos=-1;
        for(int j=i;j<sz;j++)
        {
            if(b[j]>b[i]) { pos=j; break; }
        }
        if(pos==-1) { Ans=(Ans+ans[i+1])%mod; break; }
        Ans=(Ans+(b[pos]-b[i])*ans[i+1]%mod)%mod;
        i=pos;
    }

    printf("%lld\n",Ans);
    return 0;
}

 

CodeForces 689E Mike and Geometry Problem

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原文地址:http://www.cnblogs.com/zufezzt/p/5875975.html

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