10.4 考试

T1

是一道语文题

随便sort一下就行了

 

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <cmath>
#include <algorithm>
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
const ll N=10006;
const ll mod=((ll)1<<31)-1;

ll n;
ll a[N],b[N],c[N];
ll num[N];
ll con;

void lisan()
{
    con=0;
    sort(b+1,b+1+n);
    b[0]=-3132;

    for(int i=1;i<=n;++i)
        if(b[i]!=b[i-1])
          c[++con]=b[i];

    int t1,t2;
    for(int i=1;i<=con;++i)
    {
        t1=(i-1==0?con:i-1);
        t2=(i==con?1:i+1);
        if( (c[t1]+c[i])%mod==c[t2] )
            num[i]=1;
        else
            num[i]=0;
    }
}

int er(int l,int r,ll vv)
{
    int mid;
    while(l<=r)
    {
        mid=(l+r)>>1;
        if(c[mid]==vv)
            return mid;
        if(c[mid]<vv)
            l=mid+1;
        else
            r=mid-1;
    }
}

int main(){

    //freopen("T1.in","r",stdin);

    int T=0;

    while(~scanf("%lld",&n))
    {
        ++T;
        for(int i=1;i<=n;++i)
        {
            scanf("%lld",&a[i]);
            b[i]=a[i];
        }
        //printf("n=%d\n",n);
        lisan();
        if(con==1)
        {
            printf("Case #%d: -1\n",T);
            continue;
        }
        int ans=0,temp;
        for(int i=1;i<=n;++i)
        {
            temp=er(1,con,a[i]);
            ans+=num[temp];
        }
        printf("Case #%d: %d\n",T,ans);
    }

}

 

 

 

T2

根据 f[i]=(f[i]+n-i+1)%i 递推即可

想法是每次T人后，重新标号，根据这个逆推

 

#pragma GCC optimize("O3")
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <cmath>
#include <algorithm>
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
const int N=10000006;

int T;
int n;

int main(){

    //freopen("T2.in","r",stdin);

    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        int ans=0;
        for(int i=1;i<=n;++i)
            ans=(ans+n-i+1)%i;
        printf("%d\n",ans+1);
    }
/*
2
2
3
*/
}

 

 

 

T3

首先指数%phi(p)，然后phi(p)是合数，用孙子定理

Lucas求组合数，要预处理阶乘及逆元

 

 

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
const int N=1000006;
const int mod=54184622;

ll mi(ll a,int ci,int mm)
{
    ll ans=1;
    while(ci)
    {
        if(ci&1)
            ans=ans*a%mm;
        a=a*a%mm;
        ci>>=1;
    }
    return ans%mm;
}

ll gcd(ll x,ll y){return y==0?x:gcd(y,x%y);}

void egcd(ll a,ll b,ll &x,ll &y)
{
    if(b==0)
    {
        x=1;y=0;
        return ;
    }
    egcd(b,a%b,x,y);
    ll temp=x;
    x=y;
    y=temp-a/b*y;
}

int ou(int x)
{
    int tt=x;
    for(int i=2;i*i<=x;++i)
        if(x%i==0)
        {
            tt=tt-tt/i;
            while(x%i==0)
                x/=i;
        }
    if(x!=1)
        tt=tt-tt/x;
    return tt;
}

int n,g;
ll p[106]={0,2,3,5,7,129011};
ll a[106];
ll jie[7][N],jieni[7][N],ni[7][N];

ll cm(int n,int m,int order)
{
    if(n<m)
        return 0;
    return jie[order][n]*jieni[order][m]%p[order]*jieni[order][n-m]%p[order];
}

ll Lucas(int n,int m,int order)
{
    if(m==0)
        return 1;
    return cm(n%p[order],m%p[order],order)*Lucas(n/p[order],m/p[order],order)%p[order];
}

ll CRT()
{
    ll Mi,M=1,ans=0,x,y;
    for(int i=1;i<=5;++i)
        M*=p[i];
    //printf("M=%lld\n",M);
    for(int i=1;i<=5;++i)
    {
        Mi=M/p[i];
        egcd(Mi,p[i],x,y);
        ans=(ans+Mi*x*a[i]%M)%M;
        
    }
    if(ans<0)
        ans+=M;
    return ans;
}

void chu()
{
    for(int i=1;i<=5;++i)
    {
        ni[i][1]=1;
        for(int j=2;j<N;++j)
            ni[i][j]=(ll)(p[i]-p[i]/j)*ni[i][p[i]%j]%p[i];
        jieni[i][0]=jie[i][0]=1;
        for(int j=1;j<N;++j)
        {
            jie[i][j]=jie[i][j-1]*j%p[i];
            jieni[i][j]=jieni[i][j-1]*ni[i][j]%p[i];
        }
    }
}

int main(){
    
    //freopen("in.in","r",stdin);

    chu();

    scanf("%d%d",&n,&g);
    for(int i=1;i<=n;++i)
    {
        if( gcd(i,n)==1 )
            for(int j=1;j<=5;++j)
                a[j]=(a[j]+Lucas(g,i,j))%p[j];
    }

    printf("%lld",mi(n,CRT(),mod) );
}

 

 

暴露出的问题就是我的数学十分薄弱