2018焦作网络赛G-Give Candies

G: Give Candies

时间限制: 1 Sec  内存限制: 128 MB

题目描述

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.

输入

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

输出

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

样例输入

1
4

样例输出

8

分析ans=2^(n-1)

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 const int maxn=1e7+7;
 5 const ll mod=1e9+7;
 6 char arr[maxn];
 7 ll bit[15];
 8 ll qpow(ll a,ll b,ll c)
 9 {
10     if(b==0)return 1;
11     ll ans=1;
12     while(b)
13     {
14         if(b&1)ans=(ans*a)%mod;
15         a=a*a%mod;
16         b>>=1;
17     }
18     return ans%mod;
19 }
20 ll inv(ll a,ll m)
21 {
22     if(a==1)return 1;
23     return inv(m%a,m)*(m-m/a)%m;
24 }
25 int main()
26 {
27     int t;
28     scanf("%d",&t);
29     bit[0]=1;
30     ll inv2=inv(2,mod);
31     for(int i=1;i<10;i++)
32     {
33         bit[i]=bit[i-1]*2;
34     }
35     while(t--)
36     {
37         scanf("%s",arr);
38         int len=strlen(arr);
39         ll ans=1;
40         for(int i=0;i<len;i++)
41         {
42             ans=qpow(ans,10,mod)%mod;
43             ans=ans*bit[arr[i]-‘0‘]%mod;
44         }
45         ans*=inv2;
46         printf("%lld\n",(ans+mod)%mod);
47     }
48     return 0;
49 }

原文地址：https://www.cnblogs.com/CharlieWade/p/11293918.html