ACM-ICPC 2018 焦作赛区网络预赛G Give Candies

浅浅的花香味﹌ 2022-05-12 07:56 143阅读 0赞

## 题目：[点击打开链接][Link 1] ##

## 题意：给你n个东西，叫你把n分成任意段，这样的分法有几种。 ##

## 分析：（HDU 4704原题）隔板法，ans=C(1,n-1)+C(2,n-1)+...+C(n-1,n-1)=2^(n-1)。需要用到欧拉降幂公式，参考[打开][Link 2]，证明需要用到欧拉定理，**a^(φ(m))同余1(mod m) (a与m互质)，参考[打开][Link 3]。** ##

## 代码： ##

#pragma GCC optimize(2)
    #pragma GCC optimize(3)
    #pragma GCC optimize(4)
    #pragma comment(linker, "/STACK:102400000,102400000")
    #include<unordered_map>
    #include<unordered_set>
    #include<algorithm>
    #include<iostream>
    #include<fstream>
    #include<complex>
    #include<cstdlib>
    #include<cstring>
    #include<cassert>
    #include<iomanip>
    #include<string>
    #include<cstdio>
    #include<bitset>
    #include<vector>
    #include<cctype>
    #include<cmath>
    #include<ctime>
    #include<stack>
    #include<queue>
    #include<deque>
    #include<list>
    #include<set>
    #include<map>
    using namespace std;
    #define pt(a) cout<<a<<endl
    #define debug test
    #define mst(ss,b) memset((ss),(b),sizeof(ss))
    #define rep(i,a,n) for (int i=a;i<=n;i++)
    #define per(i,a,n) for (int i=n-1;i>=a;i--)
    #define all(x) (x).begin(),(x).end()
    #define fi first
    #define se second
    #define SZ(x) ((int)(x).size())
    #define ll long long
    #define ull unsigned long long
    #define pb push_back
    #define mp make_pair
    #define inf 0x3f3f3f3f
    #define eps 1e-10
    #define PI acos(-1.0)
    typedef pair<int,int> PII;
    const ll mod = 1e9+7;
    const int N = 1e6+10;
    
    ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
    ll qp(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
    int to[4][2]={
       {-1,0},{1,0},{0,-1},{0,1}};
    
    int t;
    string s;
    
    int main() {
        ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
        cin>>t;
        while(t--) {
            cin>>s;
            int l=s.size();
            ll res=0,phi=mod-1;
            rep(i,0,l-1) res=(res*10+s[i]-'0')%phi;
            cout<<qp(2,res-1+phi)<<endl;
        }
        return 0;
    }

[Link 1]: https://nanti.jisuanke.com/t/31716
[Link 2]: https://blog.csdn.net/weixin_38686780/article/details/81272848
[Link 3]: https://blog.csdn.net/hillgong/article/details/4214327