Give Candies
時間限制: 1 Sec 內存限制: 128 MB
提交: 243 解決: 92
[提交] [狀態] [命題人:admin]
題目描述
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.

輸入
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

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

樣例輸入
復制樣例數據
1
4
樣例輸出
8

題目大意：
有$n$塊餅干，將其分給$n$個人，問有多少種分法，順序不同算兩種分法。
例如：有三塊餅干，則分法為：
1+1+1
2+1
1+2
3
共4種

解題思路：
通過列舉前幾項，很容易發現，若有$n$塊餅干，則有$2^{n?1}$種分法，由于$n$的范圍很大，所以需知道一個定理，即: $a^b=a^{\phi (b)}$
而對于質數$1000000007$來說，$\phi (1000000007)=1000000007?1$
所以可以先將指數縮小，再使用快速冪即可

代碼：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <set>
#include <utility>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define inf 0x3f3f3f3f
#define rep(i,l,r) for(int i=l;i<=r;i++)
#define lep(i,l,r) for(int i=l;i>=r;i--)
#define ms(arr) memset(arr,0,sizeof(arr))
//priority_queue<int,vector<int> ,greater<int> >q;
const int maxn = (int)1e5 + 5;
const ll mod = 1e9+7;
char s[100100];
ll quickpow(ll n,ll p) {ll base=2;ll ans=1;while(n) {if(n&1) ans=(ans*base)%p;n>>=1;base=(base*base)%p;}return ans;
}
int main() 
{#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);#endif//freopen("out.txt", "w", stdout);//ios::sync_with_stdio(0),cin.tie(0);int T;scanf("%d",&T);while(T--) {scanf("%s",s);ll nape=0;ll p=1000000007;int len=strlen(s);for(int i=0;i<len;i++) {ll x=(ll)(s[i]-'0');nape=(nape*10LL+x)%(p-1);}nape=(nape-1+p-1)%(p-1);ll ans=quickpow(nape,p);printf("%lld\n",ans);}return 0;
}