ACM-ICPC 2018 焦作赛区网络预赛 G Give Candies

There are NNN children in kindergarten. Miss Li bought them NNN 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)(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 TTT, the number of test case.

The next TTT lines, each contains an integer NNN.

1≤T≤1001 le T le 1001≤T≤100

1≤N≤101000001 le N le 10^{100000}1≤N≤10100000

Output

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

样例输入 复制

1
4

样例输出 复制

8


题意：
   给n个小孩发糖果，总共有n个糖果，小孩按顺序排好，把糖果分发，不必每个小孩都有糖果，但是如果不是每个孩子都有糖果，那么只能是在后面的小孩没有糖果。问有多少种方案。

题解：
   找到规律，对于每一个n，方案数为2^n,这里应用快速幂，但是n太大了，想办法减小它，这里要应用一下费马小定理（假如p是质数，且gcd(a,p)=1，那么 a(p-1)≡1(mod p))，即（a^n）%mod,如果mod为质数，a^(n%(mod-1))%mod。（本题中mod为1e9+7）


代码：

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
const ll mod=1e9+7;

ll Pow(ll a,ll b)
{
    ll ans=1;
    a%=mod;
    while(b){
        if(b&1) ans=ans*a%mod;
        a=a*a%mod;
        b=b>>1;
    }
    return ans;
}

int main()
{
    int t;
    ll i,n;
    char a[100100];
    cin>>t;
    while(t--){
        scanf("%s",a);
        int len=strlen(a);
        n=0;
        for(i=0;i<len;i++){
            n=(n*10+(a[i]-'0'))%(mod-1);
        }
        printf("%lld
",Pow(2,n-1)%mod);
    }
    return 0;
}