NOIP模拟测试30「return·one·magic」

magic

题解

首先原式指数肯定会爆$long$ $long$

首先根据欧拉定理我们可以将原式换成$N^{\sum\limits_{i=1}^{i<=N} [gcd(i,N)==1] C_{G}^{i} \%phi(p)}\%p$

根据欧拉函数是积性的得出$phi(54184622)=phi(2)*phi(27092311)$

然后$phi(27092311)=27092310$ $phi(2)=1$所以$phi(54184622)=27092310$

于是我们现在要求的就是$N^{\sum\limits_{i=1}^{i<=N} [gcd(i,N)==1] C_{G}^{i} \%27092310}\%p$

$27092310=2*3*5*7*129011$然后裸的$crt$求组合数板子求就完了

注意:你要预处理出阶乘和逆元,否则会超时

代码

#include<bits/stdc++.h>
#define ll long long
#define A 333333
ll k,p,n,g;
//phi(54184622)=27092310
//27092310=2*3*5*7*129011
ll w[7]={0,2,3,5,7,129011,54184622},jie[6][A],ni[6][A],dl[A],b[A];
ll exgcd(ll a,ll b,ll &x,ll &y){if(b==0){x=1;y=0;return a;}ll gcd=exgcd(b,a%b,x,y);ll t=x;x=y;y=t-a/b*y;return gcd;
}
ll meng(ll x,ll k,ll cix){ll ans=1;for(;k;k>>=1,x=x*x%w[cix])if(k&1)ans=ans*x%w[cix];return ans;
}
ll china(){ll x,y,a=0,m,n=1;for(ll i=1;i<=5;i++)n*=w[i];for(ll i=1;i<=5;i++){m=n/w[i];exgcd(w[i],m,x,y);a=(a+y*m*b[i])%n;}if(a>0) return a;return a+n;
}
ll gcd(ll x,ll y){if(y==0) return x;return gcd(y,x%y);
}
ll jic(ll n,ll m,ll cix){if(m>n) return 0;if(m==0) return 1;return jie[cix][n]%w[cix]*ni[cix][n-m]%w[cix]*ni[cix][m]%w[cix];
}
ll lucas(ll n,ll m,ll cix){if(n==0)return 1;return jic(n%w[cix],m%w[cix],cix)*lucas(n/w[cix],m/w[cix],cix)%w[cix];
}
using namespace std;
int main()
{scanf("%lld%lld",&n,&g);    for(ll i=1;i<=min(g,n);i++){if(gcd(i,n)==1)dl[++dl[0]]=i;}for(ll i=1;i<=5;i++){jie[i][0]=1;ni[i][0]=1;for(ll j=1;j<w[i];j++)jie[i][j]=jie[i][j-1]*j%w[i];ni[i][w[i]-1]=meng(jie[i][w[i]-1],w[i]-2,i);for(ll j=w[i]-2;j>=1;j--)ni[i][j]=ni[i][j+1]*(j+1)%w[i];for(ll j=1;j<=dl[0];j++)(b[i]+=lucas(g,dl[j],i))%=w[i];}ll j=china();ll k=meng(n,j,6);cout<<k<<endl;//模w「i」 剩余b「i」    
}

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one

题解

美妙的约瑟夫问题,

范围特别大考虑线性推

然而我懒的说了

代码特别简单,只是上文稍做修改

代码

#include<bits/stdc++.h>
using namespace std;
#define ll int
#define A 1000000
ll ans,t,n;
int main(){scanf("%d",&t);while(t--){scanf("%d",&n);ans=0;for(ll i=n-1;i>=0;i--)ans=(ans+i)%(n-i+1);printf("%d\n",ans+1);}
}