数学 FZU 2074 Number of methods

 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 }