BZOJ3034: Heaven Cow与God Bull

来源：互联网发布：数据库服务器编辑：程序博客网时间：2024/06/04 05:31
根据欧拉函数的表达式我们发现要使答案最大则这个一定有若干个质数相乘
#include<cstdio>#include<iostream>#include<cstring>#include<algorithm>using namespace std;#define ll long longconst ll maxn=1000000;ll prime[maxn+1],tot;bool check[maxn+1];const    ll Bt=7;const ll M=10000000;struct H{    ll len;    ll A[6001];    H(){len=0;memset(A,0,sizeof(A));}    void get()    {        char c[26001];        scanf("%s",c);        ll l=strlen(c);        for(ll i=0;i<l;i++)c[i]-='0';        len=l/Bt+(l%Bt?1:0);        ll t=l%Bt;        if(l%Bt)        {            for(ll i=l%Bt;i;i--)                A[len-1]=A[len-1]*10+c[l%Bt-i];        }        else            A[len-1]=c[6]+10*(c[5]+c[4]*10+c[3]*100+c[2]*1000+c[1]*10000+c[0]*100000);        for(ll i=0;i<len-1;i++)            l=(len-i-1)*Bt+(t==0?Bt:t),A[i]=c[l-1]+c[l-2]*10+c[l-3]*100+c[l-4]*1000+c[l-5]*10000+c[l-6]*100000+c[l-7]*1000000;    }    ll &operator [](ll x){return A[x];}    H operator *(ll a){        H c;        for(ll j=0;j<len;j++)c[j]=A[j]*a;        c.len=len;        for(ll i=0;i<c.len||c[i];i++)        {            c[i+1]+=c[i]/M;            c[i]%=M;            c.len=max(c.len,i+1);        }        return c;    }    void print()    {        printf("%lld",A[len-1]);        for(ll i=len-2;~i;i--)            printf("%07lld",A[i]);    }};bool cmp(H a,H b){       if(a.len^b.len)return a.len<b.len;    for(ll i=a.len-1;~i;i--)        if(a[i]^b[i])return a[i]<b[i];    return true;}struct Q{ll no;H A,ans;bool operator <(Q a){return no<a.no;}};bool cmp2(Q a,Q b){return cmp(a.A,b.A);}bool cmp3(Q a,Q b){return a.no<b.no;}Q S[201];char c;int main(){//  while(true)//  /{//      H A;//      A.get();    //  A.print();//  }    for(ll i=2;i<=maxn;i++)    {        if(!check[i])prime[++tot]=i;        ll k,j;        for(k=prime[j=1]*i;k<=maxn;k=prime[++j]*i)        {            check[k]=true;            if(i%prime[j]==0)break;        }    }    ll n;    scanf("%lld",&n);    for(ll i=1;i<=n;i++)    {        S[i].A.get();        S[i].no=i;    }    sort(S+1,S+1+n,cmp2);    H last,no;    last.len=1;    last[0]=1;    no=last;    ll pos=1;    for(ll i=1;i<=n;i++)    {        while(cmp(no,S[i].A))last=no,no=no*prime[pos++];        S[i].ans=last;    }    sort(S+1,S+1+n,cmp3);    for(ll i=1;i<=n;i++)        S[i].ans.print(),putchar('\n');    return 0;}
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