大数的幂

问题 G: Give Candies

时间限制: 1 Sec  内存限制: 128 MB
提交: 258  解决: 101
[提交] [状态] [命题人:admin]

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

复制样例数据

1
4

样例输出

8

思路：这题只要计算出2的n-1次方即可

费马小定理 : 如果mod为质数,（a^n )%mod =  a^(n%(mod-1)) %mod

#include <iostream>
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod=1e9+7;
const int maxn = 1e6+10;
char  a[maxn];
ll qpow(ll a,ll b)
{
    ll ret=1;
    while(b)
    {
        if (b&1) ret=ret*a%mod;
        a=a*a%mod;
        b>>=1;
    }
    return ret;
}
int main()
{
    int T,len;
    cin>>T;
    while(T--)
    {
        ll sum=0;
        scanf("%s",a);
        len=strlen(a);
        for (int i=0;i<len;i++)
        {
            sum=(sum*10+a[i]-48)%(mod-1);
        }
        if (sum==0) sum=mod-1;
        ll ans=qpow(2,sum-1);
        printf("%lld
",ans);
    }
    return 0;
}

其实这个题也可以使用欧拉降幂，这个公式其实是费马小定理的推广，费马小定理是欧拉定理的一种特殊情况。

欧拉函数即是1-n与n互质的数的个数，指数的欧拉函数当然为本身-1

欧拉降幂公式：

(a^b)%m=a^(b%phi(m))%m

贴一份队友的代码

#include<bits/stdc++.h>
using namespace std;
 
typedef long long ll;
const ll mod=1e9+7;
string s;
 
inline ll phi(int n)
{
    int ans=n,temp=n;
    for (int i=2; i*i<=temp;i++)
    {
        if (temp%i==0)
        {
            ans-=ans/i;
            while(temp%i==0) temp/=i;
        }
    }
    if (temp>1) ans-=ans/temp;
    return ans;
}
inline ll qpow(ll x,ll n,ll mod)
{
    ll ans=1;
    while(n)
    {
        if (n&1) ans=ans*x%mod;
        x=x*x%mod;
        n/=2;
    }
    return ans;
}
 
int main()
{
    int T;
    ll phic=phi(mod);
    cin>>T;
    while(T--)
    {
        cin>>s;
        int i,len=s.size();
        ll res=0,ans;
        for (i=0;i<len;i++)
        {
            res=res*10+s[i]-48;
            if (res>phic)
                break;
        }
        if (i==len)
            ans=qpow(2,res-1,mod)%mod;
        else
        {
            res=0;
            for (int i=0;i<len;i++)
            {
                res=res*10+s[i]-48;
                res%=phic;
            }
            ans=qpow(2,res+phic-1,mod)%mod;
        }
        cout<<ans<<endl;
    }
    return 0;
}