数学 FZU 2074 Number of methods

 

题目传送门

 1 /*  2  数学：假设取了第i个，有C(n-1)(i-1)种取法  3  则ans = sum (C(n-1)(i-1)) (1<i<=n) 即2^(n-1)  4 */  5 #include <cstdio>  6 #include <algorithm>  7 #include <cstring>  8 #include <cmath>  9 using namespace std; 10 11 typedef long long ll; 12 const int MAXN = 1e4 + 10; 13 const int INF = 0x3f3f3f3f; 14 const int MOD = 1000000007; 15 16 int main(void) //FZU 2074 Number of methods 17 int n; 18 while (scanf ("%d", &n) == 1) 19  { 20 ll sum = 1; 21 for (int i=1; i<n; ++i) 22  { 23 sum = sum * 2 % MOD; 24  } 25 printf ("%I64d\n", sum); 26  } 27 28 return 0; 29 }