Python(2.7.3) 解法, 执行用时: 20ms, 内存消耗: 3044K, 提交时间: 2020-06-06 14:58:29

def fastExpMod(b, e, m):
    result = 1
    while e != 0:
        if (e&1) == 1:
            result = (result * b) % m
        e >>= 1
        b = (b*b) % m
    return result

m = int (input())
while m:
    n = int (input())
    cnt=n*fastExpMod(2,n-1,1000000007)%1000000007
    print(cnt)
    m = m-1;

C++11(clang++ 3.9) 解法, 执行用时: 3ms, 内存消耗: 492K, 提交时间: 2020-06-06 15:48:06

#include<bits/stdc++.h>
using namespace std;
const int MOD=1e9+7;
typedef long long ll;

int t;
ll n;

ll quick(ll a,ll b)
{
	ll res=1;
	while(b) {
		if(b&1) res=res*a%MOD;
		a=a*a%MOD;
		b>>=1;
	}
	return res;
}

int main()
{
	cin>>t;
	while(t--){
		cin>>n;
		cout<<n*quick(2,n-1)%MOD<<endl;
	}
}

C++14(g++5.4) 解法, 执行用时: 4ms, 内存消耗: 580K, 提交时间: 2020-06-06 14:00:37

#include<bits/stdc++.h>
using namespace std;
const int MOD=1e9+7;
int t;
typedef long long ll;
ll n;
ll quick(ll a,ll b)
{
	ll res=1;
	while(b)
	{
		if(b&1) res=res*a%MOD;
		a=a*a%MOD;
		b>>=1;
	}
	return res;
}
int main()
{
	cin>>t;
	while(t--)
	{
		cin>>n;
		cout<<n*quick(2,n-1)%MOD<<endl;
	}
}

Python3(3.9) 解法, 执行用时: 21ms, 内存消耗: 2808K, 提交时间: 2021-03-23 15:26:10

md = 1000000000 + 7
for _ in range(int(input())):
    n = int(input())
    ans = pow(2,n - 1,md)
    ans *= n
    ans %= md
    print(int(ans))