ACM-ICPC 2018 焦作赛区网络预赛- G：Give Candies（费马小定理，快速幂）

There are N 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), 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 NNN.

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

题目来源

ACM-ICPC 2018 焦作赛区网络预赛

题意

计算2^(n-1)的值

思路

利用费马小定理： $a^{b}modc=a^{b*mod{(c-1)}}modc$ 求出(n-1)%(1e9+7)的值，然后利用快速幂计算结果

注意：用int型会TLE，要用long long

#include <bits/stdc++.h>

#define ms(a) memset(a,0,sizeof(a))

#define ll long long

const ll maxn=1e6+10;

const int mod=1e9+7;

using namespace std;

char ch[maxn];

ll Pow(ll a, ll b)

{

	ll res=1;

	while(b)

	{

		if(b&1)

			res=res*a%mod;

		b>>=1;

		a=a*a%mod;

	}

	return res;

}

int main(int argc, char const *argv[])

{

	ios::sync_with_stdio(false);

	register int t;

	ll ans,l;

	cin>>t;

	while(t--)

	{

		ans=0;

		cin>>ch;

		l=strlen(ch);

		for(register int i=0;i<l;i++)

		{

			ans=ans*10+ch[i]-'0';

			ans%=(mod-1);

		}

		cout<<Pow(2,ans-1)<<endl;

	}

	return 0;

}

巴特西

ACM-ICPC 2018 焦作赛区网络预赛- G：Give Candies（费马小定理，快速幂）

Input

Output

题意

思路

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