ACM训练联盟周赛 Teemo's formula

Teemo has a formula and he want to calculate it quickly.

The formula is $\sum_{i=1}^{n}k^{2}C_{n}^{k}$ .

As the result may be very large, please output the result mod 1000000007.

Input Format

The input contains several test cases, and the first line is a positive integer T indicating the number of test cases which is up to 10^5.

For each test case, the first line contains an integer n(1<=n<=10^9).

Output Format

For each test case, output a line containing an integer that indicates the answer.

样例输入

2
2
3

样例输出

6
24

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstdlib>
 5 #include <cstring>
 6 #include <string>
 7 #include <deque>
 8 #include <map>
 9 #include <vector>
10 #include <stack>
11 using namespace std;
12 #define  ll long long 
13 #define  N   29
14 #define  M 1000000000
15 #define  gep(i,a,b)  for(int  i=a;i<=b;i++)
16 #define  gepp(i,a,b) for(int  i=a;i>=b;i--)
17 #define  gep1(i,a,b)  for(ll i=a;i<=b;i++)
18 #define  gepp1(i,a,b) for(ll i=a;i>=b;i--)    
19 #define  mem(a,b)  memset(a,b,sizeof(a))
20 #define  ph  push_back
21 #define  mod  1000000007
22 ll poww(ll a,ll b){//pow会编译错误
23     ll ans=1%mod;
24     while(b){
25         if(b&1)  {
26             ans=ans*a%mod;
27         }
28         b>>=1;
29         a=a*a%mod;
30     }
31     return ans%mod;
32 }
33 int t;
34 ll n;
35 /*
36 i从1到n  i*i*C(n,i)的累加和
37 
38 在N个人里面选若干人，再选一个正司令、副司令 的方法数目
39 两个司令可以是同一个人
40 1 ： 同一人 C(n,1)*2^(n-1)//一定至少有一个人是司令，其他的N-1个人（每人有两种可能性）可以是，也可以不是
41 2 :  两个人 A(n,2)*2^(n-2) 同理 ，2个人要考虑顺序
42 也就是 n*(n+1)*2^(n-2) 
43 */
44 
45 int main()
46 {
47     scanf("%d",&t);
48     while(t--){
49         scanf("%lld",&n);
50         if(n==1) {printf("1\n");continue;}//特判
51         ll ans=n*(n+1)%mod*poww(2,n-2)%mod;
52         printf("%lld\n",ans);
53     }
54     return 0;
55 }