codeforces 689E E. Mike and Geometry Problem(组合数学)
题目链接:
3 seconds
256 megabytes
standard input
standard output
Mike wants to prepare for IMO but he doesn't know geometry, so his teacher gave him an interesting geometry problem. Let's definef([l, r]) = r - l + 1 to be the number of integer points in the segment [l, r] with l ≤ r (say that ). You are given two integers nand k and n closed intervals [li, ri] on OX axis and you have to find:
In other words, you should find the sum of the number of integer points in the intersection of any k of the segments.
As the answer may be very large, output it modulo 1000000007 (109 + 7).
Mike can't solve this problem so he needs your help. You will help him, won't you?
The first line contains two integers n and k (1 ≤ k ≤ n ≤ 200 000) — the number of segments and the number of segments in intersection groups respectively.
Then n lines follow, the i-th line contains two integers li, ri ( - 109 ≤ li ≤ ri ≤ 109), describing i-th segment bounds.
Print one integer number — the answer to Mike's problem modulo 1000000007 (109 + 7) in the only line.
3 2
1 2
1 3
2 3
5
3 3
1 3
1 3
1 3
3
3 1
1 2
2 3
3 4
6 题意: 在n个区间里选k个,得到的f等于区间交的点数;求所有的选择方案的和; 思路: 对于每个点可以发现,当这个点被num个线段覆盖时,这个点就会被选C(num,k)次,ans=∑C(num,k);
但是区间很大,点的数目居多,所以不可能一个点一个点的这样算,可以发现,相邻的点如果被相同数目的线段覆盖,那么这些点就可以合并成一个区间,所以ans=∑len*C(num,k),len表示这个区间点的个数;看这个点被覆盖了多少次可以采用跟树状数组那样的方法,左右端点+-1; AC代码:
//#include <bits/stdc++.h>
#include <vector>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <cstring>
#include <algorithm>
#include <cstdio> using namespace std;
#define For(i,j,n) for(int i=j;i<=n;i++)
#define Riep(n) for(int i=1;i<=n;i++)
#define Riop(n) for(int i=0;i<n;i++)
#define Rjep(n) for(int j=1;j<=n;j++)
#define Rjop(n) for(int j=0;j<n;j++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef long long LL;
template<class T> void read(T&num) {
char CH; bool F=false;
for(CH=getchar();CH<''||CH>'';F= CH=='-',CH=getchar());
for(num=;CH>=''&&CH<='';num=num*+CH-'',CH=getchar());
F && (num=-num);
}
int stk[], tp;
template<class T> inline void print(T p) {
if(!p) { puts(""); return; }
while(p) stk[++ tp] = p%, p/=;
while(tp) putchar(stk[tp--] + '');
putchar('\n');
} const LL mod=1e9+;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=2e5+;
const int maxn=;
const double eps=1e-; int n,k,l[N],r[N];
LL dp[N]; map<int,int>mp; LL pow_mod(int x,LL y)
{
LL s=,base=(LL)x;
while(y)
{
if(y&)s=s*base%mod;
base=base*base%mod;
y>>=;
}
return s;
} void Init()
{
dp[k]=;
For(i,k+,N)
{
LL x=i,temp=pow_mod(x-k,mod-);
dp[i]=dp[i-]*x%mod*temp%mod;
}
}
vector<int>ve;
int main()
{
read(n);read(k);
Init();
For(i,,n)
{
read(l[i]);
mp[l[i]-]++;
ve.push_back(l[i]-);
read(r[i]);
mp[r[i]]--;
ve.push_back(r[i]);
}
sort(ve.begin(),ve.end());
LL ans=;
int num=,prepo=-1e9-;
int w=ve.size();
for(int i=;i<w;i++)
{
int tempo=ve[i],len=tempo-prepo;
if(num>=k)ans=ans+dp[num]*(LL)len%mod,ans%=mod;
if(prepo!=tempo) prepo=tempo,num+=mp[tempo];
}
cout<<ans<<"\n";
return ;
}
codeforces 689E E. Mike and Geometry Problem(组合数学)的更多相关文章
- codeforces 689 E. Mike and Geometry Problem 组合数学 优先队列
给定一个函数: f([l,r]) = r - l + 1; f(空集) = 0; 即f函数表示闭区间[l,r]的整点的个数 现在给出n个闭区间,和一个数k 从n个区间里面拿出k个区间,然后对这k个区间 ...
- codeforces 361 E - Mike and Geometry Problem
原题: Description Mike wants to prepare for IMO but he doesn't know geometry, so his teacher gave him ...
- CodeForces 689E Mike and Geometry Problem (离散化+组合数)
Mike and Geometry Problem 题目链接: http://acm.hust.edu.cn/vjudge/contest/121333#problem/I Description M ...
- Codeforces Round #361 (Div. 2) E. Mike and Geometry Problem 离散化 排列组合
E. Mike and Geometry Problem 题目连接: http://www.codeforces.com/contest/689/problem/E Description Mike ...
- Codeforces Round #361 (Div. 2) E. Mike and Geometry Problem 【逆元求组合数 && 离散化】
任意门:http://codeforces.com/contest/689/problem/E E. Mike and Geometry Problem time limit per test 3 s ...
- Codeforces Round #361 (Div. 2) E. Mike and Geometry Problem 离散化+逆元
E. Mike and Geometry Problem time limit per test 3 seconds memory limit per test 256 megabytes input ...
- CodeForces 689E Mike and Geometry Problem
离散化,树状数组,组合数学. 这题的大致思路和$HDU$ $5700$一样.都是求区间交的问题.可以用树状数组维护一下. 这题的话只要计算每一个$i$被统计了几次,假设第$i$点被统计了$ans[i] ...
- 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 ...
- 【codeforces 798C】Mike and gcd problem
[题目链接]:http://codeforces.com/contest/798/problem/C [题意] 给你n个数字; 要求你进行若干次操作; 每次操作对第i和第i+1个位置的数字进行; 将 ...
随机推荐
- python和搜索
# -*- coding: UTF-8 -*- import re # 搜索逻辑 def querylogic(list): query = {} if len(list) > 1 or len ...
- 按Esc按钮关闭layer弹窗
//按Esc关闭弹出框 $(document).ready(function () { }).keydown( function (e) { if (e.which === 27) { layer. ...
- 【思维+贪心】codeforces Game of the Rows
http://codeforces.com/contest/839/problem/B [题意] 给定n组人,告诉每组人的人数,这些人要在飞机上坐座位 飞机上座位的分布看图可以知道,12 3456 ...
- Vim command handbook
/* 本篇文章已经默认你通过了vimtuor训练并能熟练使用大部分命令.此篇文章主要是对于tutor命令的总结和梳理.适合边学习边记忆 tutor那个完全是在学习中记忆 符合认知规律但是练习有限.所以 ...
- Hankson 的趣味题(codevs 1172)
题目描述 Description Hanks 博士是BT (Bio-Tech,生物技术) 领域的知名专家,他的儿子名叫Hankson.现在,刚刚放学回家的Hankson 正在思考一个有趣的问题.今天在 ...
- 《TCP/IP详解卷1:协议》——第3章 IP:网际协议(转载)
1.引言 IP是TCP/IP协议族中最核心的协议.所有的TCP.UDP.ICMP及IGMP数据都以IP数据报格式传输.IP提供不可靠. 无连接的数据报传送服务. (1)不可靠 它不能保证IP数据报能成 ...
- django学习之- 数据缓存
5种配置:开发调试 CACHES = { 'default': { 'BACKEND': 'django.core.cache.backends.dummy.DummyCache', # 引擎内存CA ...
- Codeforces 659E New Reform【DFS】
题目链接: http://codeforces.com/problemset/problem/659/E 题意: 给定n个点和m条双向边,将双向边改为单向边,问无法到达的顶点最少有多少个? 分析: 无 ...
- $.post()用法例子
1:删除用户操作 $('.delete').click(function(){ var classid=$(this).parent().siblings().eq(0).children().val ...
- hdu 3943
数位dp #include <cstdio> #include <cstdlib> #include <cmath> #include <stack> ...