Gym 101775A - Chat Group - [简单数学题][2017EC-Final]
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题目链接:http://codeforces.com/gym/101775/problem/A
It is said that a dormitory with 6 persons has 7 chat groups ^_^. But the number can be even larger: since every 3 or more persons could make a chat group, there can be 42 different chat groups.
Given N persons in a dormitory, and every K or more persons could make a chat group, how many different chat groups could there be?
Input
The input starts with one line containing exactly one integer T which is the number of test cases.
Each test case contains one line with two integers N and K indicating the number of persons in a dormitory and the minimum number of persons that could make a chat group.
1?≤?T?≤?100.
1?≤?N?≤?10^9.
3?≤?K?≤?10^5.
Output
For each test case, output one line containing "Case #x: y" where x is the test case number (starting from 1) and y is the number of different chat groups modulo 1000000007.
Example
Input
1
6 3
Output
Case #1: 42
题意:
听说一个寝室六个人有七个群?但实际上如果六人寝里三个人及以上组成不同的群的话,可以组成 $42$ 个群……
现在给出一个 $n$ 人寝室,要求计算 $k$ 人及以上的不同的群可以建几个?
题解:
$C_{n}^{k}+ cdots + C_{n}^{n} = (C_{n}^{0}+ C_{n}^{1} + cdots + C_{n}^{n}) - (C_{n}^{0}+ C_{n}^{1} + cdots + C_{n}^{k-1})$
又根据二项式展开可知 $2^n = (1+1)^{n} = C_{n}^{0} imes 1^{0} imes 1^{n} + C_{n}^{1} imes 1^{1} imes 1^{n-1} + cdots + C_{n}^{n} imes 1^{n} imes 1^{0} = C_{n}^{0} + C_{n}^{1} + cdots + C_{n}^{n}$
因此答案即为 $2^{n} - (C_{n}^{0}+ C_{n}^{1} + cdots + C_{n}^{k-1})$。
运用累乘的方式计算 $C_{n}^{0}, C_{n}^{1}, cdots, C_{n}^{k-1}$,注意除法要使用逆元。
AC代码:
#include<bits/stdc++.h> using namespace std; typedef long long ll; const ll MOD=1000000007; ll n,k; ll fpow(ll a,ll b) { ll r=1,base=a%MOD; while(b) { if(b&1) r*=base,r%=MOD; base*=base; base%=MOD; b>>=1; } return r; } ll inv(ll a){return fpow(a,MOD-2);} int main() { int T; cin>>T; for(int kase=1;kase<=T;kase++) { scanf("%lld%lld",&n,&k); if(n<k) { printf("Case #%d: 0 ",kase); continue; } ll sum=1+n,tmp=n; for(ll i=1;i<=k-2;i++) { tmp=(((tmp*(n-i))%MOD)*inv(i+1))%MOD; sum=(sum+tmp)%MOD; } ll ans=(fpow(2,n)-sum+MOD)%MOD; printf("Case #%d: %d ",kase,ans); } }
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