Information
- ID
- 1544
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 5
- Tags
- # Submissions
- 24
- Accepted
- 14
- Uploaded By
#include<bits/stdc++.h> using namespace std;
typedef long long ll;
ll mod = 1000000007;
ll quick_pow(ll a, ll b, ll p) {//快速幂求2^n-1
ll ans = 1;
while (b) {
if (b & 1) ans = (ans * a) % p;
a = (a * a) % p;
b >>= 1;
}
return ans;
}//考察了一点二项式定理
int main() {//答案为1cn1+2cn2+3cn3+....ncnn,合并成ans下面的公式
ll n;
cin >> n;
ll ans = n*quick_pow(2, n - 1, mod)%mod;//ans=n*2^n-1
cout << ans;
return 0;
}
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