神秘钥匙

看到题就得以下公式：

ans = 1*C(n,1) + 2*C(n,2) + 3*C(n,3) + ... + n*C(n,n)
    = n*2^(n-1)

计算过程：

S1  = 0*C(n,0) + 1*C(n,1) + 2*C(n,2) + ... + n*C(n,n)

S2  = n*C(n,0) + (n-1)*C(n,1) + (n-2)*C(n,2) + ... + 0*C(n,n)

S   = S1 + S2 = n*[C(n,0)+C(n,1)+...+C(n,n)]

    = n*2^n

∵ S1 = S2 (利用 C(n,m) = C(n,n-m))

∴ 2*S1 = n*2^n

    S1 = n*2^(n-1)

代码：

#include <iostream>
#include <cstdio>
using namespace std;
typedef long long ll;
const int mod = 1000000007;

ll  qpow(ll a, ll n){
    ll  ans = 1;
    while(n){
        if(n&1)
            ans = (a*ans) % mod;
        a = a*a % mod;
        n >>= 1;
    }
    return ans;
}

int main() {
    ll n;
    scanf("%lld",&n);
    ll ans = n*qpow(2,(n-1))%mod;
    printf("%lld\n",ans);
    return 0;
}