【规律】Gym 100739L Many recursions

给出a，求递归式g(k)的初始条件g(0);

可以看出来g(a) = 1，从后往前推。写个模拟程序可以看出来其实g(0) = 2^a，那么就是一个简单地快速幂取模问题了。

#include <cstdio>
#include <iostream>
#include <sstream>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <string>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <algorithm>
using namespace std;
#define ll unsigned long long
#define _cle(m, a) memset(m, a, sizeof(m))
#define repu(i, a, b) for(int i = a; i < b; i++)
#define repd(i, a, b) for(int i = b; i >= a; i--)
#define sfi(n) scanf("%d", &n)
#define pfi(n) printf("%d
", n)
#define MAXN 4010
const int N = 20005;
#define mod 1000000007

ll modexp(ll a, ll b) //(a^b)%mod; 
{
    ll ret = 1;
    ll tmp = a;
    while(b)
    {
        if(b&0x1) ret = ret*tmp%mod;
        tmp = tmp*tmp%mod;
        b >>= 1;
    }
    return ret;
}

int main()
{
    ll a;
    while(~scanf("%I64d", &a))
    {
        ll ans = modexp((ll)2, a);
        printf("%I64d
", ans);
    }
    return 0;
}