2018ACM-ICPC焦作赛区网络赛题目---Give Candies

                                                  Give Candies

• 时间限制： 1000ms         内存限制： 65536K

There are N children in kindergarten. Miss Li bought them N 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), 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 T, the number of test case.

The next T lines, each contains an integer N.

1≤T≤100

1≤N≤10^100000

Output

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

样例输入

1
4

样例输出

8

思路：题目经过分析，答案就为计算2^(n-1)%mod的值即可，当n很大时，最大为10^100000,求2^n%mod   费马小定理求解，即有 2^n%mod=(2^(n%(mod-1))%mod; 

AC代码：

//当n很大时，最大为10^100000,求2^n%mod   费马小定理求解
//即有 2^n%mod=(2^(n%(mod-1))%mod; 
#include
#include
#include
using namespace std;
typedef long long ll;
const ll mod=1000000007;
string s;
ll a[100010];
ll mod_pow(ll x,ll n){
    if(n==0) return 1;
    ll res=mod_pow(x*x%mod,n/2);
    if(n&1) res=res*x%mod;
    return res;
}
ll fun(ll k){
	ll sum=0;
	for(ll i=1;i<=k;i++){
		sum=(sum*10+a[i])%(mod-1);
	}
	return sum;
}
int main(){
	int t;
	cin>>t;
	while(t--){
		cin>>s;
		ll k=s.size();
		for(ll i=0;i