Teams
2013-09-22 13:32
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http://acm.hust.edu.cn/vjudge/contest/view.action?cid=31675#problem/H
先从N个人中选一个班长,然后在选队员;
显然选班长有N种情况,接下来就是选队员,就是从剩下来的人中间选C( N - 1 , 1 ) + C(N - 1 , 2 ) + 。。。。+ C( N - 1 , N - 1 ) ;
根据二项式定理可以得到( 1 + 1 ) ^ ( N - 1 )
所以总的情况根据乘法原理得N × ( 2 ^ ( N - 1 ) ;
先从N个人中选一个班长,然后在选队员;
显然选班长有N种情况,接下来就是选队员,就是从剩下来的人中间选C( N - 1 , 1 ) + C(N - 1 , 2 ) + 。。。。+ C( N - 1 , N - 1 ) ;
根据二项式定理可以得到( 1 + 1 ) ^ ( N - 1 )
所以总的情况根据乘法原理得N × ( 2 ^ ( N - 1 ) ;
// File Name: uva11609.cpp // Author: bo_jwolf // Created Time: 2013年09月22日 星期日 13:17:07 #include<vector> #include<list> #include<map> #include<set> #include<deque> #include<stack> #include<bitset> #include<algorithm> #include<functional> #include<numeric> #include<utility> #include<sstream> #include<iostream> #include<iomanip> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<ctime> using namespace std; typedef unsigned long long int INT ; const INT Mod = 1000000007 ; INT Pow_Mod( INT a , INT b ) { INT ans = 1 ; while( b ) { if( b & 1 ) { ans = ( ans * a ) % Mod ; b-- ; } a = ( a * a ) % Mod ; b /= 2 ; } return ans ; } int main() { int Case , n , temp = 1 ; scanf( "%d" , &Case ) ; while( Case-- ) { scanf( "%d" , &n ) ; printf( "Case #%d: %llu\n" , temp++ , ( ( n % Mod ) * ( Pow_Mod( 2 , n - 1 ) % Mod ) ) % Mod ) ; } return 0; }
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