Codeforces Round #361 (Div. 2) E. Mike and Geometry Problem

题目链接:传送门

题目大意:给你n个区间，求任意k个区间交所包含点的数目之和。

题目思路:将n个区间都离散化掉，然后对于一个覆盖的区间，如果覆盖数cnt>=k，则数目应该加上 区间长度*（cnt与k的组合数） ans=ans+(len*C(cnt,k))%mod;

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <stack>
#include <cctype>
#include <queue>
#include <string>
#include <vector>
#include <set>
#include <map>
#include <climits>
#define lson root<<1,l,mid
#define rson root<<1|1,mid+1,r
#define fi first
#define se second
#define ping(x,y) ((x-y)*(x-y))
#define mst(x,y) memset(x,y,sizeof(x))
#define mcp(x,y) memcpy(x,y,sizeof(y))
using namespace std;
#define gamma 0.5772156649015328606065120
#define MOD 1000000007
#define inf 0x3f3f3f3f
#define N 200005
#define maxn 1050
typedef pair<int,int> PII;
typedef long long LL;

int n,k,m,cnt;
LL fac[N];
struct Node{
    int x,v;
    bool operator<(const Node&a)const{
        if(x==a.x)return v>a.v;
        return x<a.x;
    }
}node[N<<1];
int a[N<<1],sum[N<<1];
int has[N<<1];
LL ksm(LL a,LL b){
    LL res=1;
    while(b){
        if(b&1)res=res*a%MOD;
        b>>=1;
        a=a*a%MOD;
    }
    return res;
}
LL C(LL n,LL m){
    if(n<m||m<0)return 0;
    LL s1=fac[n],s2=fac[n-m]*fac[m]%MOD;
    return s1*ksm(s2,MOD-2)%MOD;
}
int main(){
    int i,j,group,x,y,v;
    fac[0]=1;cnt=0;
    for(i=1;i<N;++i)fac[i]=fac[i-1]*i%MOD;
    scanf("%d%d",&n,&k);
    for(i=1;i<=n;++i){
        scanf("%d%d",&x,&y);
        node[cnt].x=x;node[cnt++].v=1;
        node[cnt].x=y+1;node[cnt++].v=-1;
    }
    sort(node,node+cnt);
    LL ans=0,la,num=0; ///la是区间左端点
    for(i=0;i<cnt;++i){
        if(num>=k) ans=(ans+(node[i].x-la*1ll)*C(num,k))%MOD;
        la=node[i].x;
        num+=node[i].v;
    }
    printf("%I64d
",ans);
    return 0;
}