Give Candies

•  1000ms
•  65536K

There are NN children in kindergarten. Miss Li bought them NN candies. To make the process more interesting, Miss Li comes up with the rule: All the children line up according to their student number (1...N)(1...N), and each time a child is invited, Miss Li randomly gives him some candies (at least one). The process goes on until there is no candy. Miss Li wants to know how many possible different distribution results are there.

Input

The first line contains an integer TT, the number of test case.

The next TT lines, each contains an integer NN.

1 \le T \le 1001≤T≤100

1 \le N \le 10^{100000}1≤N≤10100000

Output

For each test case output the number of possible results (mod 1000000007).

樣例輸入複製

1
4

樣例輸出複製

8

題目來源

思路：

這道題很簡單，首先我們輕易就看出了答案就是2^(n - 1)，但是由於n非常大，直接快速冪肯定是不行的，於是可以進位來求

很明顯2^(10n) = (2^n)^10，由於冪次是n - 1，所以不能最後盲目的除以2，可以利用費馬小定理離線求出逆元，於是就解出來了

程式碼：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll mod = (ll)1e9 + 7;
const ll maxn = (ll)1e5 + 10;
const ll inv = 500000004;
char num[maxn];
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()
{
    int t;
    scanf("%d",&t);
    while (t --)
    {
        scanf("%s",num);
        int len = strlen(num);
        ll ans = 1;
        for (int i = 0;i < len;i ++)
        {
            ans = quick(ans,10);
            ans = ans * quick(2,num[i] - '0') % mod;
        }
        ans = ans * inv % mod;
        printf("%lld\n",ans);
    }
    return 0;
}