ACM-ICPC 2018 焦作赛区网络预赛 G. Give Candies

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).

样例输入复制

1
4

样例输出复制

8

题目来源

ACM-ICPC 2018 焦作赛区网络预赛

通过打表可知，答案是2^n-1 ，但是n又太大，直接计算会超时。所以这里就需要用到欧拉函数了。话不多说上代码：

#include<iostream>
#include<cstring>
#include<string>
#include<algorithm>
using namespace std;
char a[1005][1005];
int main()
{
    int n;
    a[0][0] = '1';
    int before = 1;
    for(int i = 1;i <= 100; ++i)
    {
        int k = 0;
        int mod = 0;
        for(int j = 0;j < before;++j)
        {
            a[i][k++] = ((a[i - 1][j] - '0') * 2 + mod) % 10+48;
            mod = (((a[i - 1][j] - '0') * 2) + mod) / 10;
        }
        if(mod)
            a[i][k++] = mod + 48;
        before=k;
    }
    while(cin>>n)
    {
        int len=strlen(a[n]);
        for(int i=len-1;i>=0;--i)
            cout<<a[n][i];
        cout<<endl;
    }
}