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  • [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
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  • 原文地址:https://www.cnblogs.com/hate13/p/4466046.html
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