zoukankan      html  css  js  c++  java
  • [scu 4423] Necklace

    4423: Necklace

    Description

    
    baihacker bought a necklace for his wife on their wedding anniversary.
    A necklace with N pearls can be treated as a circle with N points where the
    distance between any two adjacent points is the same. His wife wants to color
    every point, but there are at most 2 kinds of color. How many different ways
    to color the necklace. Two ways are said to be the same iff we rotate one
    and obtain the other.

    Input

    
    The first line is an integer T that stands for the number of test cases.
    Then T line follow and each line is a test case consisted of an integer N.
    
    Constraints:
    T is in the range of [0, 4000]
    N is in the range of [1, 1000000000]
    N is in the range of [1, 1000000], for at least 75% cases.

    Output

    
    For each case output the answer modulo 1000000007 in a single line.

    Sample Input

    
    6
    1
    2
    3
    4
    5
    20

    Sample Output

    
    2
    3
    4
    6
    8
    52488

    Author

    
    baihacker

    疯狂地模板题,受不了,比赛的时候连这个定理都没听过,还傻乎乎地想了好久,晕死- -

    #include<iostream>
    #include<cstdio>
    #include<cmath>
    #include<cstring>
    using namespace std;
    #define ll long long
    #define N 32000
    
    ll tot;
    ll prime[N+10];
    bool isprime[N+10];
    ll phi[N+10];
    void init()
    {
        memset(phi,-1,sizeof(phi));
        memset(isprime,1,sizeof(isprime));
        tot=0;
        phi[1]=1;
        isprime[0]=isprime[1]=0;
        for(ll i=2;i<=N;i++)
        {
            if(isprime[i])
            {
                prime[tot++]=i;
                phi[i]=i-1;
            }
            for(ll j=0;j<tot;j++)
            {
                if(i*prime[j]>N) break;
                isprime[i*prime[j]]=0;
                if(i%prime[j]==0)
                {
                    phi[i*prime[j]]=phi[i]*prime[j];
                    break;
                }
                else
                    phi[i*prime[j]]=phi[i]*(prime[j]-1);
            }
        }
    }
    ll euler(ll n)
    {
        if(n<=N) return phi[n];
        ll ret=n;
        for(ll i=0;prime[i]*prime[i]<=n;i++)
        {
            if(n%prime[i]==0)
            {
                ret-=ret/prime[i];
                while(n%prime[i]==0) n/=prime[i];
            }
        }
        if(n>1) ret-=ret/n;
        return ret;
    }
    ll quickpow(ll a,ll b,ll MOD)
    {
        a%=MOD;
        ll ret=1;
        while(b)
        {
            if(b&1) ret=(ret*a)%MOD;
            a=(a*a)%MOD;
            b>>=1;
        }
        return ret;
    }
    ll exgcd(ll a,ll b,ll& x, ll& y)
    {
        if(b==0)
        {
            x=1;
            y=0;
            return a;
        }
        ll d=exgcd(b,a%b,y,x);
        y-=a/b*x;
        return d;
    }
    ll inv(ll a,ll MOD)
    {
        ll x,y;
        exgcd(a,MOD,x,y);
        x=(x%MOD+MOD)%MOD;
        return x;
    }
    void solve(ll n,ll MOD)
    {
        ll i,t1,t2,ans=0;
        for(i=1;i*i<=n;i++)
        {
            if(n%i==0)
            {
                if(i*i!=n)
                {
                    t1=euler(n/i)%MOD*quickpow(2,i,MOD);
                    t2=euler(i)%MOD*quickpow(2,n/i,MOD);
                    ans=(ans+t1+t2)%MOD;
                }
                else
                    ans=(ans+euler(i)*quickpow(2,i,MOD))%MOD;
            }
        }
        ans=ans*inv(n,MOD)%MOD;
        printf("%d
    ",ans);
    }
    int main()
    {
        init();
        ll T,n;
        ll MOD=1000000007;
        scanf("%lld",&T);
        while(T--)
        {
            scanf("%lld",&n);
            solve(n,MOD);
        }
        return 0;
    }
     
    趁着还有梦想、将AC进行到底~~~by 452181625
  • 相关阅读:
    HDU 5650 异或
    HDU 5646
    HDU 5645
    P2075 [NOIP2012T5]借教室 区间更新+二分查找
    HDU 5641
    读写分离
    linux执行cmd之一
    html2image
    挂载引起的权限问题
    如何防止sql注入
  • 原文地址:https://www.cnblogs.com/hate13/p/4466046.html
Copyright © 2011-2022 走看看